Rs Aggarwal 2021 2022 for Class 10 Maths Chapter 7 - Triangles

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Rs Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 7 Triangles are provided here with simple step-by-step explanations. These solutions for Triangles are extremely popular among Class 10 students for Maths Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 10 Maths Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 372:

Answer:

(i)
$In △ ABC, it is given that DE ∥ BC .$
Applying Thales' theorem, we get:
$\frac{A D}{D B} = \frac{A E}{E C}$

$∵$ AD = 3.6 cm , AB = 10 cm, AE = 4.5 cm
$∴$ DB = 10 $-$ 3.6 = 6.4 cm
$o r, \frac{3.6}{6.4} = \frac{4.5}{E C} o r, E C = \frac{6.4 \times 4.5}{3.6} o r, E C = 8 c m Thus, A C = A E + E C = 4.5 + 8 = 12.5 c m$

(ii)

$In △ ABC, it is given that DE ∥ BC . Applying Thales' Theorem, we get : \frac{AD}{DB} = \frac{AE}{EC} Adding 1 to both sides, we get : \frac{AD}{DB} + 1 = \frac{AE}{EC} + 1 \Rightarrow \frac{AB}{DB} = \frac{AC}{EC} \Rightarrow \frac{13.3}{DB} = \frac{11.9}{5.1} \Rightarrow DB = \frac{13.3 \times 5.1}{11.9} = 5.7 cm$
$Therefore, A D = A B - D B = 13.5 - 5.7 = 7.6 cm$

(iii)
$In △ ABC, it is given that DE ∥ BC . Applying Thales' theorem, we get : \frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{4}{7} = \frac{AE}{EC} Adding 1 to both the sides, we get : \frac{11}{7} = \frac{AC}{EC} \Rightarrow EC = \frac{6.6 \times 7}{11} = 4.2 cm Therefore, AE = AC - EC = 6.6 - 4.2 = 2.4 cm$

(iv)

$In △ ABC, it is given that DE ∥ BC . Applying Thales' theorem, we get : \frac{AD}{AB} = \frac{AE}{AC} \Rightarrow \frac{8}{15} = \frac{AE}{AE + EC} \Rightarrow \frac{8}{15} = \frac{AE}{AE + 3.5} \Rightarrow 8 AE + 28 = 15 AE \Rightarrow 7 AE = 28 \Rightarrow AE = 4 cm$

Page No 372:

Answer:

(i)
$In △ ABC, it is given that DE ∥ BC . Applying Thales' theorem, we have : \frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{x}{x - 2} = \frac{x + 2}{x - 1} \Rightarrow x (x - 1) = (x - 2) (x + 2) \Rightarrow x^{2} - x = x^{2} - 4 \Rightarrow x = 4 cm$

(ii)
$In △ ABC, it is given that DE ∥ BC . Applying Thales' theorem, we have : \frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{4}{x - 4} = \frac{8}{3 x - 19} \Rightarrow 4 (3 x - 19) = 8 (x - 4) \Rightarrow 12 x - 76 = 8 x - 32 \Rightarrow 4 x = 44 \Rightarrow x = 11 cm$

(iii)
$In △ ABC, it is given that DE ∥ BC . Applying Thales' theorem, we have : \frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{7 x - 4}{3 x + 4} = \frac{5 x - 2}{3 x} \Rightarrow 3 x (7 x - 4) = (5 x - 2) (3 x + 4) \Rightarrow 21 x^{2} - 12 x = 15 x^{2} + 14 x - 8 \Rightarrow 6 x^{2} - 26 x + 8 = 0 \Rightarrow (x - 4) (6 x - 2) = 0 \Rightarrow x = 4, \frac{1}{3} ∵ x \neq \frac{1}{3} (as if x = \frac{1}{3} then A E will become negative) ∴ x = 4 cm$

Page No 372:

Answer:

(i) We have:

$\frac{AD}{DB} = \frac{5.7}{9.5} = 0.6 cm \frac{AE}{EC} = \frac{4.8}{8} = 0.6 cm Hence, \frac{AD}{DB} = \frac{AE}{EC} Applying the converse of Thales' theorem, we conclude that DE ∥ BC .$

(ii)
We have:
AB = 11.7 cm, DB = 6.5 cm
Therefore,
AD = 11.7 $-$ 6.5 = 5.2 cm
Similarly,
AC = 11.2 cm, AE = 4.2 cm
Therefore,
EC = 11.2 $-$ 4.2 = 7 cm

$Now, \frac{AD}{DB} = \frac{5.2}{6.5} = \frac{4}{5} \frac{AE}{EC} = \frac{4.2}{7} Thus, \frac{AD}{DB} \neq \frac{AE}{EC}$

Applying the converse of Thales' theorem,
we conclude that DE is not parallel to BC.

(iii)
We have:
AB = 10.8 cm, AD = 6.3 cm
Therefore,
DB = 10.8 $-$ 6.3 = 4.5 cm
Similarly,
AC = 9.6 cm, EC = 4 cm
Therefore,
AE = 9.6 $-$ 4 = 5.6 cm
Now,
$\frac{AD}{DB} = \frac{6.3}{4.5} = \frac{7}{5} \frac{AE}{EC} = \frac{5.6}{4} = \frac{7}{5} \Rightarrow \frac{AD}{DB} = \frac{AE}{EC} Applying the converse of Thales' theorem, we conclude that DE ∥ BC .$

(iv)
We have:
AD = 7.2 cm, AB = 12 cm
Therefore,
DB = 12 $-$ 7.2 = 4.8 cm
Similarly,
AE = 6.4 cm, AC = 10 cm
Therefore,
EC = 10 $-$ 6.4 = 3.6 cm
Now,
$\frac{AD}{DB} = \frac{7.2}{4.8} = \frac{3}{2} \frac{AE}{EC} = \frac{6.4}{3.6} = \frac{16}{9} Thus, \frac{AD}{DB} \neq \frac{AE}{EC} Applying the converse of Thales' theorem, we conclude that DE is not parallel to BC .$

Page No 372:

Answer:

(i)
$It is given that AD bisects ∠ A . Applying angle - bisector theorem in △ ABC, we get : \frac{BD}{DC} = \frac{AB}{AC} \Rightarrow \frac{5.6}{DC} = \frac{6.4}{8} \Rightarrow DC = \frac{8 \times 5.6}{6.4} = 7 cm$

(ii)
$It is given that AD bisects ∠ A . Applying angle - bisector theorem in △ ABC, we get :$

$\frac{BD}{DC} = \frac{AB}{AC} Let BD be x cm . Therefore, DC = (6 - x) cm \Rightarrow \frac{x}{6 - x} = \frac{10}{14} \Rightarrow 14 x = 60 - 10 x \Rightarrow 24 x = 60 \Rightarrow x = 2.5 cm Thus, BD = 2.5 cm DC = 6 - 2.5 = 3.5 cm$

(iii)
$It is given that AD bisects ∠ A . Applying angle - bisector theorem in △ ABC, we get :$

$\frac{BD}{DC} = \frac{AB}{AC} BD = 3.2 cm, BC = 6 cm Therefore, DC = 6 - 3.2 = 2.8 cm \Rightarrow \frac{3.2}{2.8} = \frac{5.6}{AC}$

$\Rightarrow A C = \frac{5.6 \times 2.8}{3.2} = 4.9 cm$

(iv)
$It is given that AD bisects ∠ A . Applying angle - bisector theorem in △ ABC, we get :$

$\frac{B D}{D C} = \frac{A B}{A C} \Rightarrow \frac{B D}{3} = \frac{5.6}{4} \Rightarrow B D = \frac{5.6 \times 3}{4} \Rightarrow B D = 4.2 cm$

Hence, BC = 3 + 4.2 = 7.2 cm

Page No 373:

Answer:

Given: ABCD is a parallelogram
To prove:
$(i) \frac{DM}{MN} = \frac{DC}{BN} (ii) \frac{DN}{DM} = \frac{AN}{DC}$
Proof: In △DMC and △NMB
$∠$ DMC = $∠$ NMB (Vertically opposite angle)
$∠$ DCM = $∠$ NBM (Alternate angles)
By AAA- similarity
△DMC ~ △NMB
$∴ \frac{DM}{MN} = \frac{DC}{BN}$
$Now, \frac{MN}{DM} = \frac{BN}{DC}$
Adding 1 to both sides, we get
$\frac{MN}{DM} + 1 = \frac{BN}{DC} + 1 \Rightarrow \frac{MN + DM}{DM} = \frac{BN + DC}{DC} \Rightarrow \frac{MN + DM}{DM} = \frac{BN + AB}{DC} [∵ ABCD is a parallelogram] \Rightarrow \frac{DN}{DM} = \frac{AN}{DC}$

Page No 373:

Answer:

Let the trapezium be ABCD with E and F as the mid points of AD and BC, respectively.
Produce AD and BC to meet at P.

In $△ P A B, D C ∥ A B .$
Applying Thales' theorem, we get:
$\frac{P D}{D A} = \frac{P C}{C B} Now, E and F are the midpoints of A D and B C, respectively . \Rightarrow \frac{P D}{2 D E} = \frac{P C}{2 C F} \Rightarrow \frac{P D}{D E} = \frac{P C}{C F} Applying the converse of Thales' theorem in ∆ P E F, we get that D C ∥ E F . Hence, E F ∥ A B .$

Thus. EF is parallel to both AB and DC.
This completes the proof.

Page No 373:

Answer:

In trapezium ABCD, $A B ∥ C D$ and the diagonals AC and BD intersect at O.
Therefore,

$\frac{A O}{O C} = \frac{B O}{O D} \Rightarrow \frac{2 x + 1}{5 x - 7} = \frac{7 x + 1}{7 x - 5} \Rightarrow (5 x - 7) (7 x + 1) = (7 x - 5) (2 x + 1) \Rightarrow 35 x^{2} + 5 x - 49 x - 7 = 14 x^{2} - 10 x + 7 x - 5 \Rightarrow 21 x^{2} - 41 x - 2 = 0 \Rightarrow 21 x^{2} - 42 x + x - 2 = 0 \Rightarrow 21 x (x - 2) + 1 (x - 2) = 0 \Rightarrow (x - 2) (21 x + 1) = 0 \Rightarrow x = 2, - \frac{1}{21} ∵ x \neq - \frac{1}{21} ∴ x = 2$

Page No 373:

Answer:

In △ABC, ∠B = ∠C
∴AB = AC (Sides opposite to equal angle are equal)
Subtracting BM from both sides, we get
AB − BM = AC − BM
⇒AB − BM = AC − CN (∵BM = CN)
⇒AM = AN
∴∠AMN = ∠ANM (Angles opposite to equal sides are equal)
Now, In △ABC,
∠A + ∠B + ∠C = 180^o               .....(1)
                                                               (Angle Sum Property of triangle)
Again In In △AMN,
∠A +∠AMN + ∠ANM = 180^o    ......(2)
                                                               (Angle Sum Property of triangle)
From (1) and (2), we get
∠B + ∠C = ∠AMN + ∠ANM
⇒2∠B = 2∠AMN
⇒∠B = ∠AMN
Since, ∠B and ∠AMN are corresponding angles.
∴ MN∥BC

Page No 373:

Answer:

$In △ C A B, P Q ∥ A B . Applying Thales' theorem, we get : \frac{C P}{P B} = \frac{C Q}{Q A} . . . (1)$
Similarly, applying Thales' theorem in $△ B D C, where P R ∥ B D$ , we get:
$\frac{C P}{P B} = \frac{C R}{R D} . . . (2) Hence, from (1) and (2), we have : \frac{C Q}{Q A} = \frac{C R}{R D}$

Applying the converse of Thales' theorem, we conclude that $Q R ∥ A D i n △ A D C$ .
This completes the proof.

Page No 373:

Answer:

Join BX and CX.
It is given that BC is bisected at D.
∴ BD = DC
It is also given that OD = OX
The diagonals OX and BC of quadrilateral BOCX bisect each other.
Therefore, BOCX is a parallelogram.
∴ BO $∥ C X and B X ∥ C O$
$B X ∥ C F and C X ∥ B E$
$B X ∥ O F$ and $C X ∥ O E$
Applying Thales' theorem in $∆$ ABX, we get:
$\frac{A O}{A X} = \frac{A F}{A B} . . . (1) Also, in △ A C X, C X ∥ O E . Therefore by Thales' theorem, we get : \frac{A O}{A X} = \frac{A E}{A C} . . . (2)$
From (1) and (2), we have:
$\frac{A F}{A B} = \frac{A E}{A C}$
Applying the converse of Thales' theorem in $△ A B C, E F ∥ C B$ .
This completes the proof.

Page No 373:

Answer:

Join DB.
We know that the diagonals of a parallelogram bisect each other.
Therefore,
CS = $\frac{1}{2}$ AC ...(i)
Also, it is given that CQ = $\frac{1}{4}$ AC ...(ii)
Dividing equation (ii) by (i), we get:
$\frac{C Q}{C S} = \frac{\frac{1}{4} A C}{\frac{1}{2} A C}$
or, CQ = $\frac{1}{2}$ CS
Hence, Q is the midpoint of CS.
Therefore, according to midpoint theorem in $△ C S D$
$P Q ∥ D S$
$if P Q ∥ D S, we can say that Q R ∥ S B$

In $△ C S B, Q is midpoint of C S and Q R ∥ S B$ .
Applying converse of midpoint theorem ,we conclude that R is the midpoint of CB.
This completes the proof.

Page No 373:

Answer:

Given:
AD = AE ...(i)
AB = AC ...(ii)
Subtracting AD from both sides, we get:
$\Rightarrow$ AB $-$ AD = AC $-$ AD
$\Rightarrow$ AB $-$ AD = AC $-$ AE (Since, AD = AE)
$\Rightarrow$ BD = EC ...(iii)
Dividing equation (i) by equation (iii), we get:

$\frac{A D}{D B} = \frac{A E}{E C}$

Applying the converse of Thales' theorem, DE $∥$ BC

$\Rightarrow ∠ D E C + ∠ E C B = 180 ° (Sum of interior angles on the same side of a transversal line is 180 ° .) \Rightarrow ∠ D E C + ∠ C B D = 180 ° (Since, A B = A C \Rightarrow ∠ B = ∠ C)$

Hence, quadrilateral BCED is cyclic.

Therefore, B,C,E and D are concyclic points.

Page No 374:

Answer:

In triangle BQP, BR bisects angle B.
Applying angle bisector theorem, we get:
$\frac{Q R}{P R} = \frac{B Q}{B P}$
$\Rightarrow B P \times Q R = B Q \times P R$
This completes the proof.

Page No 399:

Answer:

(i)
We have:
$∠ B A C = ∠ P Q R = 50 ° ∠ A B C = ∠ Q P R = 60 ° ∠ A C B = ∠ P R Q = 70 °$

Therefore, by AAA similarity theorem, $△ A B C ~ Q P R$

(ii)
We have:

$\frac{A B}{D F} = \frac{3}{6} = \frac{1}{2} a n d \frac{B C}{D E} = \frac{4.5}{9} = \frac{1}{2}$

But, $∠ A B C \neq ∠ E D F$ (Included angles are not equal)
Thus, this does not satisfy SAS similarity theorem.
Hence, the triangles are not similar.

(iii)
We have:
$\frac{C A}{Q R} = \frac{8}{6} = \frac{4}{3} a n d \frac{C B}{P Q} = \frac{6}{4.5} = \frac{4}{3} \Rightarrow \frac{C A}{Q R} = \frac{C B}{P Q}$
Also, $∠ A C B = ∠ P Q R = 80 °$
Therefore, by SAS similarity theorem, $△ A C B ~ △ R Q P$ .

(iv)
We have
$\frac{D E}{Q R} = \frac{2.5}{5} = \frac{1}{2} \frac{E F}{P Q} = \frac{2}{4} = \frac{1}{2} \frac{D F}{P R} = \frac{3}{6} = \frac{1}{2} \Rightarrow \frac{D E}{Q R} = \frac{E F}{P Q} = \frac{D F}{P R}$

Therefore, by SSS similarity theorem, $△ F E D ~ △ P Q R$

(v)
$In △ ABC ∠ A + ∠ B + ∠ C = 180 ° (Angle Sum Property) \Rightarrow 80 ° + ∠ B + 70 ° = 180 ° \Rightarrow ∠ B = 30 °$
$∠ A = ∠ M and ∠ B = ∠ N$
Therefore, by AA similarity theorem, $△ ABC ~ △ MNR$

Page No 400:

Answer:

(i)
It is given that DB is a straight line.
Therefore,
$∠ D O C + ∠ C O B = 180 ° ∠ D O C = 180 ° - 115 ° = 65 °$

(ii)
In $△ D O C$ , we have:
$∠ O D C + ∠ D C O + ∠ D O C = 180 ° Therefore, 70 ° + ∠ D C O + 65 ° = 180 ° \Rightarrow ∠ D C O = 180 - 70 - 65 = 45 °$

(iii)
It is given that $△ O D C ~ △ O B A$
Therefore,
$∠ O A B = ∠ O C D = 45 °$

(iv)
Again, $△ O D C ~ △ O B A$
Therefore,
$∠ O B A = ∠ O D C = 70 °$

Page No 400:

Answer:

(i) Let OA be x cm.
$∵$ $△ O A B ~ △ O C D$
$∴ \frac{O A}{O C} = \frac{A B}{C D} \Rightarrow \frac{x}{3.5} = \frac{8}{5} \Rightarrow x = \frac{8 \times 3.5}{5} = 5.6$
Hence, OA = 5.6 cm

(ii) Let OD be y cm

$∵$ $△ O A B ~ △ O C D$
$∴ \frac{A B}{C D} = \frac{O B}{O D} \Rightarrow \frac{8}{5} = \frac{6.4}{y} \Rightarrow y = \frac{6.4 \times 5}{8} = 4$
Hence, DO = 4 cm

Page No 401:

Answer:

Given:
$∠$ ADE = $∠$ ABC and $∠ A = ∠ A$
Let DE be x cm
Therefore, by AA similarity theorem, $△ A D E ~ △ A B C$

$\Rightarrow \frac{A D}{A B} = \frac{D E}{B C} \Rightarrow \frac{3.8}{3.6 + 2.1} = \frac{x}{4.2} \Rightarrow x = \frac{3.8 \times 4.2}{5.7} = 2.8$

Hence, DE = 2.8 cm

Page No 401:

Answer:

It is given that triangles ABC and PQR are similar.
Therefore,

$\frac{Perimeter (△ A B C)}{Perimeter (△ P Q R)} = \frac{A B}{P Q} \Rightarrow \frac{32}{24} = \frac{A B}{12} \Rightarrow A B = \frac{32 \times 12}{24} = 16 cm$

Page No 401:

Answer:

$It is given that △ ABC ~ △ DEF .$
Therefore, their corresponding sides will be proportional.
Also, the ratio of the perimeters of similar triangles is same as the ratio of their corresponding sides.

$\Rightarrow \frac{P e r i m e t e r o f △ A B C}{P e r i m e t e r o f △ D E F} = \frac{B C}{E F}$

Let the perimeter of ∆ABC be x cm.

Therefore,
$\frac{x}{25} = \frac{9.1}{6.5} \Rightarrow x = \frac{9.1 \times 25}{6.5} = 35$

Thus, the perimeter of ∆ABC is 35 cm.

Page No 401:

Answer:

$In △ B D A and △ B A C, we have : ∠ B D A = ∠ B A C = 90 ° ∠ D B A = ∠ C B A (Common) Therefore, by AA similarity theorem, △ B D A ~ △ B A C$

$\Rightarrow \frac{A D}{A C} = \frac{A B}{B C}$

$\Rightarrow \frac{A D}{0.75} = \frac{1}{1.25} \Rightarrow A D = \frac{0.75}{1.25} = 0.6 m or 60 cm$

Page No 401:

Answer:

It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.

$In △ B D C and △ A B C, we have : ∠ A B C = ∠ B D C = 90 ° (given) ∠ C = ∠ C (common) By AA similarity theorem, we get : △ B D C ~ △ A B C$

$\frac{A B}{B D} = \frac{B C}{D C} \Rightarrow \frac{5.7}{3.8} = \frac{B C}{5.4} \Rightarrow B C = \frac{5.7}{3.8} \times 5.4 = 8.1$

Hence, BC = 8.1 cm

Page No 401:

Answer:

It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.

$In △ D B A and △ D C B, we have : ∠ B D A = ∠ C D B ∠ D B A = ∠ D C B = 90 ° Therefore, by AA similarity theorem, we get : △ D B A ~ △ D C B \Rightarrow \frac{B D}{C D} = \frac{A D}{B D} \Rightarrow C D = \frac{B D^{2}}{A D}$

CD = $\frac{8 \times 8}{4} = 16$ cm

Page No 401:

Answer:

We have:

$\frac{A P}{A B} = \frac{2}{6} = \frac{1}{3} and \frac{AQ}{AC} = \frac{3}{9} = \frac{1}{3} \Rightarrow \frac{A P}{A B} = \frac{AQ}{AC} In △ A P Q and △ A B C, we have : \frac{A P}{A B} = \frac{A Q}{A C} ∠ A = ∠ A Therefore, by AA similarity theorem, we get : △ A P Q ~ △ A B C H e n c e, \frac{P Q}{B C} = \frac{AQ}{AC} = \frac{1}{3} \Rightarrow \frac{P Q}{B C} = \frac{1}{3} \Rightarrow B C = 3 P Q$
This completes the proof.

Page No 402:

Answer:

We have:
$∠ A F D = ∠ E F B (Vertically Opposite angles)$
$∵$ DA $∥$ BC
$∴$ $∠ D A F = ∠ B E F$ (Alternate angles)
$△ D A F ~ △ B E F$ (AA similarity theorem)
$\Rightarrow \frac{A F}{E F} = \frac{F D}{F B}$
or, $A F \times F B = F D \times E F$
This completes the proof.

Page No 402:

Answer:

In $△ B E D a n d △ A C B$ , we have:

$∠ B E D = ∠ A C B = 90 ° ∵ ∠ B + ∠ C = 180 ° ∴ B D ∥ A C ∠ E B D = ∠ C A B (Alternate angles) Therefore, by AA similarity theorem, we get : △ B E D ~ △ A C B \Rightarrow \frac{B E}{A C} = \frac{D E}{B C} \Rightarrow \frac{B E}{D E} = \frac{A C}{B C}$
This completes the proof.

Page No 402:

Answer:

Let AB be the vertical stick and BC be its shadow.
Given:
AB = 7.5 m, BC = 5 m

Let PQ be the tower and QR be its shadow.
Given:
QR = 24 m
Let the length of PQ be x m.

In $△ A B C and △ P Q R$ , we have:
$∠ A B C = ∠ P Q R = 90 ° ∠ A C B = ∠ P R Q (Angular elevation of the Sun at the same time)$
Therefore, by AA similarity theorem, we get:
$△ A B C ~ △ P Q R$
$\Rightarrow \frac{A B}{B C} = \frac{P Q}{Q R}$

$\Rightarrow \frac{7.5}{5} = \frac{x}{24} \Rightarrow x = \frac{7.5}{5} \times 24 = 36 m$
Therefore, PQ = 36 m

Hence, the height of the tower is 36 m.

Page No 402:

Answer:

Disclaimer: It should be $△ A P C ~ △ B C Q$ instead of ∆ACP ∼ ∆BCQ
It is given that $△$ ABC is an isosceles triangle.
Therefore,
CA = CB
$\Rightarrow ∠ C A B = ∠ C B A \Rightarrow 180 ° - ∠ C A B = 180 ° - ∠ C B A \Rightarrow ∠ C A P = ∠ C B Q$

Also,
$A P \times B Q = A C^{2} \Rightarrow \frac{A P}{A C} = \frac{A C}{B Q} \Rightarrow \frac{A P}{A C} = \frac{B C}{B Q} (∵ AC = BC)$

Thus, by SAS similarity theorem, we get:
$△ A P C ~ △ B C Q$
This completes the proof.

Page No 402:

Answer:

We have:
$\frac{A C}{B D} = \frac{C B}{C E} \Rightarrow \frac{A C}{C B} = \frac{B D}{C E} \Rightarrow \frac{A C}{C B} = \frac{C D}{C E} (Since, B D = D C as ∠ 1 = ∠ 2) Also, ∠ 1 = ∠ 2 i . e, ∠ D B C = ∠ A C B Therefore, by SAS similarity theorem, we get : △ A C B ~ △ D C E$

Page No 402:

Answer:

In △ABC, P and Q are mid points of AB and AC respectively.
So, PQ|| BC, and $PQ = \frac{1}{2} BC$       .....(1)

Similarly, In △ADC, $QR = \frac{1}{2} AD = \frac{1}{2} BC$                           .....(2)
Now, In △BCD, $SR = \frac{1}{2} BC$                  .....(3)

Similarly, In △ABD, $PS = \frac{1}{2} AD = \frac{1}{2} BC$                    .....(4)
Using (1), (2), (3) and (4)
PQ = QR = SR = PS
Since, all sides are equal
Hence, PQRS is a rhombus.

Page No 402:

Answer:

Given: AB and CD are two chords
To Prove:
$(a) △ PAC ~ △ PDB (b) PA . PB = PC . PD$

Proof: In $△ PAC and △ PDB$
$∠ APC = ∠ DPB$ (Vertically Opposite angles)
$∠ CAP = ∠ BDP$ (Angles in the same segment are equal)
By AA similarity-criterion $△ PAC ~ △ PDB$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are porportional.
$∴ \frac{PA}{PD} = \frac{PC}{PB} \Rightarrow PA . PB = PC . PD$

Page No 403:

Answer:

Given: AB and CD are two chords
To Prove:
$(a) △ PAC ~ △ PDB (b) PA . PB = PC . PD$

Proof:
$∠ ABD + ∠ ACD = 180^{\circ}$            .....(1)
                                                                         (Opposite angles of a cyclic quadrilateral are supplementary)
$∠ PCA + ∠ ACD = 180^{\circ}$            ....(2)
                                                                        (Linear Pair Angles)
Using (1) and (2), we get
$∠ ABD = ∠ PCA$
$∠ A = ∠ A$    (Common)
By AA similarity-criterion $△ PAC ~ △ PDB$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
$∴ \frac{PA}{PD} = \frac{PC}{PB} \Rightarrow PA . PB = PC . PD$

Page No 403:

Answer:

We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on the both sides of the perpendicular are similar to the whole triangle and also to each other.
(a) Now using the same property in In △BDC, we get

△CQD ∼ △DQB
$\frac{CQ}{DQ} = \frac{DQ}{QB} \Rightarrow {DQ}^{2} = QB . CQ$

Now. since all the angles in quadrilateral BQDP are right angles.
Hence, BQDP is a rectangle.
So, QB = DP and DQ = PB
$∴ {DQ}^{2} = DP . CQ$

(b)

Similarly, △APD ∼ △DPB
$\frac{AP}{DP} = \frac{PD}{PB} \Rightarrow {DP}^{2} = AP . PB \Rightarrow {DP}^{2} = AP . DQ [∵ DQ = PB]$

Page No 403:

Answer:

Since, AD and PM are the medians of ΔABC and ΔPQR respectively
Therefore, BD = DC = $\frac{B C}{2}$ and QM = MR = $\frac{Q R}{2}$ ...(1)

Now,
ΔABC ~ ΔPQR

As we know, corresponding sides of similar triangles are proportional.
Thus, $\frac{A B}{P Q} = \frac{B C}{Q R} = \frac{A C}{P R}$ ...(2)

Also, $∠ A = ∠ P, ∠ B = ∠ Q and ∠ C = ∠ R$ ...(3)

From (1) and (2), we get
$\frac{A B}{P Q} = \frac{B C}{Q R} \Rightarrow \frac{A B}{P Q} = \frac{2 B D}{2 Q M} \Rightarrow \frac{A B}{P Q} = \frac{B D}{Q M} . . . (4)$

Now, in ΔABD and ΔPQM

$\frac{A B}{P Q} = \frac{B D}{Q M} (From (4)) ∠ B = ∠ Q (From (3))$

By SAS similarity,
ΔABD ~ ΔPQM

Therefore, $\frac{A B}{P Q} = \frac{B D}{Q M} = \frac{A D}{P M}$ .

Hence, $\frac{A B}{P Q} = \frac{A D}{P M}$ .

Page No 417:

Answer:

It is given that $△ A B C ~ △ D E F$ .
Therefore, ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.

$\frac{a r (△ A B C)}{a r (△ D E F)} = \frac{B C^{2}}{E F^{2}} Let B C be x cm . \Rightarrow \frac{64}{121} = \frac{x^{2}}{(15.4)^{2}} \Rightarrow x^{2} = \frac{64 \times 15.4 \times 15.4}{121} \Rightarrow x = \sqrt{\frac{64 \times 15.4 \times 15.4}{121}} = \frac{8 \times 15.4}{11} = 11.2$

Hence, BC = 11.2 cm

Page No 417:

Answer:

It is given that $△ A B C ~ △ P Q R$ .
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
$\frac{ar (△ A B C)}{ar (△ P Q R)} = \frac{B C^{2}}{Q R^{2}} \Rightarrow \frac{9}{16} = \frac{4 . 5^{2}}{Q R^{2}} \Rightarrow Q R^{2} = \frac{4.5 \times 4.5 \times 16}{9} \Rightarrow Q R = \sqrt{\frac{4.5 \times 4.5 \times 16}{9}} = \frac{4.5 \times 4}{3} = 6 cm$

Hence, QR = 6 cm

Page No 417:

Answer:

^{$Given : ar (△ A B C) = 4 ar (∆ P Q R) \frac{ar (△ A B C)}{ar (∆ P Q R)} = \frac{4}{1} ∵ ∆ A B C ~ ∆ P Q R ∴ \frac{ar (△ A B C)}{ar (∆ P Q R)} = \frac{B C^{2}}{Q R^{2}} ∴ \frac{B C^{2}}{Q R^{2}} = \frac{4}{1}$}
$\Rightarrow Q R^{2} = \frac{12^{2}}{4} \Rightarrow Q R^{2} = 36 \Rightarrow Q R = 6 cm$
Hence, QR = 6 cm

Page No 417:

Answer:

It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Let the longest side of smaller triangle be x cm.

^{$\frac{ar (Larger triangle)}{ar (Smaller triangle)} = \frac{(Longest side of larger traingle)^{2}}{(Longest side of smaller traingle)^{2}} \Rightarrow \frac{169}{121} = \frac{26^{2}}{x^{2}} \Rightarrow x = \sqrt{\frac{26 \times 26 \times 121}{169}} = 22$}

Hence, the longest side of the smaller triangle is 22 cm.

Page No 417:

Answer:

It is given that ∆ABC ∼ ∆DEF.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Let the altitude of ∆ABC be AP, drawn from A to BC to meet BC at P and the altitude of ∆DEF be DQ, drawn from D to meet EF at Q.

Then,
$\frac{a r (∆ A B C)}{a r (∆ D E F)} = \frac{A P^{2}}{D Q^{2}} \Rightarrow \frac{100}{49} = \frac{5^{2}}{D Q^{2}} \Rightarrow \frac{100}{49} = \frac{25}{D Q^{2}} \Rightarrow D Q^{2} = \frac{49 \times 25}{100} \Rightarrow D Q = \sqrt{\frac{49 \times 25}{100}} \Rightarrow D Q = 3.5 cm$

Hence, the altitude of ∆DEF is 3.5 cm

Page No 417:

Answer:

Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively.

It is given that $△ A B C ~ △ D E F$ .
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

$∴ \frac{ar (△ A B C)}{ar (△ D E F)} = \frac{(A P)^{2}}{(D Q)^{2}} \Rightarrow \frac{ar (△ A B C)}{ar (△ D E F)} = \frac{6^{2}}{9^{2}} = \frac{36}{81} = \frac{4}{9}$

Hence, the ratio of their areas is 4 : 9

Page No 418:

Answer:

It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively.

$\frac{ar (∆ A B C)}{ar (∆ P Q R)} = \frac{A P^{2}}{D Q^{2}} \Rightarrow \frac{81}{49} = \frac{6 . 3^{2}}{D Q^{2}} \Rightarrow D Q^{2} = \frac{49}{81} \times 6 . 3^{2} \Rightarrow D Q^{2} = \sqrt{\frac{49}{81} \times 6.3 \times 6.3} = 4.9 cm$

Hence, the altitude of the other triangle is 4.9 cm.

Page No 418:

Answer:

Let the two triangles be ABC and PQR with medians AM and PN, respectively.

Therefore, the ratio of areas of two similar triangles will be equal to the ratio of squares of their corresponding medians.
$∴ \frac{ar (∆ A B C)}{ar (∆ P Q R)} = \frac{A M^{2}}{P N^{2}} \Rightarrow \frac{64}{100} = \frac{5 . 6^{2}}{P N^{2}} \Rightarrow P N^{2} = \frac{64}{100} \times 5 . 6^{2} \Rightarrow P N^{2} = \sqrt{\frac{100}{64} \times 5.6 \times 5.6} = 7 cm$
Hence, the median of the larger triangle is 7 cm.

Page No 418:

Answer:

We have:
$\frac{A P}{A B} = \frac{1}{1 + 3} = \frac{1}{4} and \frac{A Q}{A C} = \frac{1.5}{1.5 + 4.5} = \frac{1.5}{6} = \frac{1}{4} \Rightarrow \frac{A P}{A B} = \frac{A Q}{A C}$

Also, $∠ A = ∠ A$
By SAS similarity , we can conclude that ∆APQ $~$ ∆ABC.

$\frac{ar (∆ A P Q)}{ar (∆ A B C)} = \frac{A P^{2}}{A B^{2}} = \frac{1^{2}}{4^{2}} = \frac{1}{16} \Rightarrow \frac{ar (∆ A P Q)}{ar (∆ A B C)} = \frac{1}{16} \Rightarrow ar (∆ A P Q) = \frac{1}{16} \times ar (∆ A B C)$

Hence proved.

Page No 418:

Answer:

It is given that DE ∥ BC

$∴ ∠ A D E = ∠ A B C (Corresponding angles) ∠ A E D = ∠ A C B (Corresponding angles)$

By AA similarity , we can conclude that $△ A D E ~ △ A B C$ .

$∴ \frac{ar (∆ A D E)}{ar (∆ A B C)} = \frac{D E^{2}}{B C^{2}} \Rightarrow \frac{15}{ar (∆ A B C)} = \frac{3^{2}}{6^{2}} \Rightarrow ar (∆ A B C) = \frac{15 \times 36}{9} = 60 {cm}^{2}$

Hence, area of triangle ABC is 60 cm².

Page No 418:

Answer:

In $△ A B C and △ A D C$ , we have:
$∠ B A C = ∠ A D C = 90 ° ∠ A C B = ∠ A C D (common)$
By AA similarity, we can conclude that $△ B A C ~ △ A D C$ .
Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their corresponding sides.

$∴ \frac{ar (∆ B A C)}{ar (∆ A D C)} = \frac{B C^{2}}{A C^{2}} \Rightarrow \frac{ar (∆ B A C)}{ar (∆ A D C)} = \frac{13^{2}}{5^{2}} = \frac{169}{25}$

Hence, the ratio of areas of both the triangles is 169 : 25

Page No 418:

Answer:

It is given that DE $∥$ BC.
$∴ ∠ A D E = ∠ A B C (Corresponding angles) ∠ A E D = ∠ A C B (Corresponding angles)$
Applying AA similarity theorem, we can conclude that $△ A D E ~ △ A B C$ .

$∴ \frac{a r (∆ A B C)}{a r (∆ A D E)} = \frac{B C^{2}}{D E^{2}} Subtracting 1 from both sides, we get : \frac{a r (∆ A B C)}{a r (∆ A D E)} - 1 = \frac{5^{2}}{3^{2}} - 1 \Rightarrow \frac{a r (∆ A B C) - a r (∆ A D E)}{a r (∆ A D E)} = \frac{25 - 9}{9} \Rightarrow \frac{a r (BCED)}{a r (∆ A D E)} = \frac{16}{9} or, \frac{a r (∆ A D E)}{a r (BCED)} = \frac{9}{16}$

Page No 418:

Answer:

It is given that D and E are midpoints of AB and AC.
Applying midpoint theorem, we can conclude that DE $∥$ BC.
Hence, by B.P.T., we get:
$\frac{A D}{A B} = \frac{A E}{A C}$
Also, $∠ A = ∠ A$ .

Applying SAS similarity theorem, we can conclude that $△ A D E ~ △ A B C$ .
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
$∴ \frac{ar (∆ A D E)}{ar (∆ A B C)} = \frac{D E^{2}}{B C^{2}} = \frac{{(\frac{1}{2} B C)}^{2}}{B C^{2}} = \frac{1}{4}$

Page No 441:

Answer:

For the given triangle to be right-angled, the sum of square of the two sides must be equal to the square of the third side.
Here, let the three sides of the triangle be a, b and c.
(i)
a = 9 cm, b = 16 cm and c = 18 cm
Then,
$a^{2} + b^{2} = 9^{2} + 16^{2} = 81 + 256 = 337 c^{2} = 19^{2} = 361 a^{2} + b^{2} \neq c^{2}$

Thus, the given triangle is not right-angled.

(ii)
a = 7 cm, b = 24 cm and c = 25 cm
Then,
$a^{2} + b^{2} = 7^{2} + 24^{2} = 49 + 576 = 625 c^{2} = 25^{2} = 625 a^{2} + b^{2} = c^{2}$
Thus, the given triangle is a right-angled.

(iii)
a = 1.4 cm, b = 4.8 cm and c = 5 cm
Then,
$a^{2} + b^{2} = (1.4)^{2} + (4.8)^{2} = 1.96 + 23.04 = 25 c^{2} = 5^{2} = 25 a^{2} + b^{2} = c^{2}$
Thus, the given triangle is right-angled.

(iv)
a = 1.6 cm, b = 3.8cm and c = 4 cm
Then,
$a^{2} + b^{2} = (1.6)^{2} + (3.8)^{2} = 2.56 + 14.44 = 17 c^{2} = 4^{2} = 16 a^{2} + b^{2} \neq c^{2}$
Thus, the given triangle is not right-angled.

(v)
p = (a $-$ 1) cm, q = 2 $\sqrt{a}$ cm and r = (a + 1) cm
Then,
$p^{2} + q^{2} = (a - 1)^{2} + (2 \sqrt{a})^{2} = a^{2} + 1 - 2 a + 4 a = a^{2} + 1 + 2 a = (a + 1)^{2} r^{2} = (a + 1)^{2} p^{2} + q^{2} = {r^{2}}^{}$
Thus, the given triangle is right-angled.

Page No 441:

Answer:

Let the man starts from point A and goes 80 m due east to B.
Then, from B, he goes 150 m due north to C.

We need to find AC.
In right-angled triangle ABC, we have:

$A C^{2} = A B^{2} + B C^{2} A C = \sqrt{80^{2} + 150^{2}} = \sqrt{6400 + 22500} = \sqrt{28900} = 170 m$

Hence, the man is 170 m away from the starting point.

Page No 441:

Answer:

Let the man starts from point D and goes 10 m due south and stops at E. He then goes 24 m due west at F.
In right $△$ DEF, we have:
DE = 10 m, EF = 24 m

$D F^{2} = E F^{2} + D E^{2} D F = \sqrt{10^{2} + 24^{2}} = \sqrt{100 + 576} = \sqrt{676} = 26 m$

Hence, the man is 26 m away from the starting point.

Page No 441:

Answer:

Let AB and AC be the ladder and height of the building.
It is given that:
AB = 13 m and AC = 12 m
We need to find the distance of the foot of the ladder from the building, i.e, BC.
In right-angled triangle ABC, we have:

$A B^{2} = A C^{2} + B C^{2} \Rightarrow B C = \sqrt{13^{2} - 12^{2}} = \sqrt{169 - 144} = \sqrt{25} = 5 m$

Hence, the distance of the foot of the ladder from the building is 5 m

Page No 441:

Answer:

Let the height of the window from the ground and the distance of the foot of the ladder from the wall be AB and BC, respectively.
We have:
AB = 20 m and BC = 15 m
Applying Pythagoras theorem in right-angled triangle ABC, we get:

$A C^{2} = A B^{2} + B C^{2} \Rightarrow A C = \sqrt{20^{2} + 15^{2}} = \sqrt{400 + 225} = \sqrt{625} = 25 m$
Hence, the length of the ladder is 25 m.

Page No 441:

Answer:

Let the two poles be DE and AB and the distance between their bases be BE.
We have:
DE = 9 m, AB = 14 m and BE = 12 m
Draw a line parallel to BE from D, meeting AB at C.
Then, DC = 12 m and AC = 5 m
We need to find AD, the distance between their tops.

Applying Pythagoras theorem in right-angled triangle ACD, we have:

$A D^{2} = A C^{2} + D C^{2} A D^{2} = 5^{2} + 12^{2} = 25 + 144 = 169 A D = \sqrt{169} = 13 m$

Hence, the distance between the tops of the two poles is 13 m.

Page No 442:

Answer:

Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let it to be fixed at A.
Now, In right triangle ABC
By using Pythagoras theorem, we have
${AB}^{2} = {BC}^{2} + {CA}^{2} \Rightarrow 24^{2} = 18^{2} + {CA}^{2} \Rightarrow {CA}^{2} = 576 - 324 \Rightarrow {CA}^{2} = 252 \Rightarrow CA = 6 \sqrt{7} m$
Hence, the stake should be driven $6 \sqrt{7}$ m far from the base of the pole.

Page No 442:

Answer:

Applying Pythagoras theorem in right-angled triangle POR, we have:
${PR}^{2} = {PO}^{2} + {OR}^{2} \Rightarrow {PR}^{2} = 6^{2} + 8^{2} = 36 + 64 = 100 \Rightarrow PR = \sqrt{100} = 10 cm$

In ∆ PQR,

${PQ}^{2} + {PR}^{2} = 24^{2} + 10^{2} = 576 + 100 = 676 and {QR}^{2} = 26^{2} = 676 ∴ {PQ}^{2} + {PR}^{2} = {QR}^{2}$

Therefore, by applying Pythagoras theorem, we can say that ∆PQR is right-angled at P.

Page No 442:

Answer:

It is given that $△ A B C$ is an isosceles triangle.
Also, AB = AC = 13 cm
Suppose the altitude from A on BC meets BC at D. Therefore, D is the midpoint of BC.
AD = 5 cm
$△ A D B a n d △ A D C$ are right-angled triangles.
Applying Pythagoras theorem, we have:

$A B^{2} = A D^{2} + B D^{2} B D^{2} = A B^{2} - A D^{2} = 13^{2} - 5^{2} B D^{2} = 169 - 25 = 144 B D = \sqrt{144} = 12$

Hence,
BC = 2(BD) = 2 $\times 12$ = 24 cm

Page No 442:

Answer:

In isosceles ∆ ABC, we have:
AB = AC = 2a units and BC = a units
Let AD be the altitude drawn from A that meets BC at D.
Then, D is the midpoint of BC.
BD = BC = $\frac{a}{2} units$
Applying Pythagoras theorem in right-angled ∆ABD, we have:

$A B^{2} = A D^{2} + B D^{2} A D^{2} = A B^{2} - B D^{2} = (2 a)^{2} - {(\frac{a}{2})}^{2} A D^{2} = 4 a^{2} - \frac{a^{2}}{4} = \frac{15 a^{2}}{4} A D = \sqrt{\frac{15 a^{2}}{4}} = \frac{a \sqrt{15}}{2} units$

Page No 442:

Answer:

Let AD, BE and CF be the altitudes of ∆ABC meeting BC, AC and AB at D, E and F, respectively.
Then, D, E and F are the midpoints of BC, AC and AB, respectively.

In right-angled ∆ABD, we have:
AB = 2a and BD = a
Applying Pythagoras theorem, we get:
$A B^{2} = A D^{2} + B D^{2} A D^{2} = A B^{2} - B D^{2} = (2 a)^{2} - a^{2} A D^{2} = 4 a^{2} - a^{2} = 3 a^{2} A D = \sqrt{3} a units$

Similarly,
BE = $a \sqrt{3}$ units and CF = $a \sqrt{3}$ units

Page No 442:

Answer:

Let ABC be the equilateral triangle with AD as an altitude from A meeting BC at D. Then, D will be the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:

$A B^{2} = A D^{2} + B D^{2} \Rightarrow A D^{2} = 12^{2} - 6^{2} (∵ B D = \frac{1}{2} B C = 6) \Rightarrow A D^{2} = 144 - 36 = 108 \Rightarrow A D = \sqrt{108} = 6 \sqrt{3} cm$

Hence, the height of the given triangle is 6 $\sqrt{3}$ cm.

Page No 442:

Answer:

Let ABCD be the rectangle with diagonals AC and BD meeting at O.
According to the question:
AB = CD = 30 cm and BC = AD = 16 cm

Applying Pythagoras theorem in right-angled triangle ABC, we get:

$A C^{2} = A B^{2} + B C^{2} = 30^{2} + 16^{2} = 900 + 256 = 1156 A C = \sqrt{1156} = 34 cm$

Diagonals of a rectangle are equal.
Therefore, AC = BD = 34 cm

Page No 442:

Answer:

Let ABCD be the rhombus with diagonals (AC = 24 cm and BD = 10 cm) meeting at O.
We know that the diagonals of a rhombus bisect each other at right angles.
Applying Pythagoras theorem in right-angled triangle AOB, we get:
$A B^{2} = A O^{2} + B O^{2} = 12^{2} + 5^{2} A B^{2} = 144 + 25 = 169 A B = \sqrt{169} = 13 cm$

Hence, the length of each side of the rhombus is 13 cm.

Page No 442:

Answer:

In right-angled triangle AEB, applying Pythagoras theorem, we have:
$A B^{2} = A E^{2} + E B^{2}$ ...(i)
In right-angled triangle AED, applying Pythagoras theorem, we have:

$A D^{2} = A E^{2} + E D^{2}$
$\Rightarrow A E^{2} = A D^{2} - E D^{2}$ ...(ii)

$Therefore, A B^{2} = A D^{2} - E D^{2} + E B^{2} (from (i) and (ii)) A B^{2} = A D^{2} - E D^{2} + (B D - D E)^{2} = A D^{2} - E D^{2} + (\frac{1}{2} B C^{} - D E)^{2} = A D^{2} - D E^{2} + \frac{1}{4} B C^{2} + D E^{2} - B C . D E = A D^{2} + \frac{1}{4} B C^{2} - B C . D E$

This completes the proof.

Page No 442:

Answer:

Given: ∠ACB = 90⁰ and CD ⊥ AB
To Prove: $\frac{{BC}^{2}}{{AC}^{2}} = \frac{BD}{AD}$
Proof:
In $△ ACB and △ CDB$
$∠ ACB = ∠ CDB = 90^{\circ}$ (Given)
$∠ ABC = ∠ CBD$ (Common)
By AA similarity-criterion $△ ACB ~ △ CDB$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
$∴ \frac{BC}{BD} = \frac{AB}{BC} \Rightarrow {BC}^{2} = BD . AB . . . . . (1)$
In $△ ACB and △ ADC$
$∠ ACB = ∠ ADC = 90^{\circ}$ (Given)
$∠ CAB = ∠ DAC$ (Common)
By AA similarity-criterion $△ ACB ~ △ ADC$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
$∴ \frac{AC}{AD} = \frac{AB}{AC} \Rightarrow {AC}^{2} = AD . AB . . . . . (2)$
Dividing (2) by (1), we get
$\frac{{BC}^{2}}{{AC}^{2}} = \frac{BD}{AD}$

Page No 442:

Answer:

(i)
In right-angled triangle AEC, applying Pythagoras theorem, we have:

$A C^{2} = A E^{2} + E C^{2} \Rightarrow b^{2} = h^{2} + {(x + \frac{a}{2})}^{2} = h^{2} + x^{2} + \frac{a^{2}}{4} + a x . . . (i) In right - angled triangle AED, we have : A D^{2} = A E^{2} + E D^{2} \Rightarrow p^{2} = h^{2} + x^{2} . . . (i i) Therefore, from (i) and (ii), b^{2} = p^{2} + a x + \frac{a^{2}}{4}$

(ii)
In right-angled triangle AEB, applying Pythagoras theorem, we have:
$A B^{2} = A E^{2} + E B^{2} \Rightarrow c^{2} = h^{2} + {(\frac{a}{2} - x)}^{2} (∵ B D = \frac{a}{2} a n d B E = B D - x) \Rightarrow c^{2} = h^{2} + x^{2} - a x + \frac{a^{2}}{4} (∵ h^{2} + x^{2} = p^{2}) \Rightarrow c^{2} = p^{2} - a x + \frac{a^{2}}{4}$

(iii)

Adding (i) and (ii), we get:
$\Rightarrow b^{2} + c^{2} = p^{2} + a x + \frac{a^{2}}{4} + p^{2} - a x + \frac{a^{2}}{4} = 2 p^{2} + a x - a x + \frac{a^{2} + a^{2}}{4} = 2 p^{2} + \frac{a^{2}}{2}$

(iv)
Subtracting (ii) from (i), we get:
$b^{2} - c^{2} = p^{2} + a x + \frac{a^{2}}{4} - (p^{2} - a x + \frac{a^{2}}{4}) = p^{2} - p^{2} + a x + a x + \frac{a^{2}}{4} - \frac{a^{2}}{4} = 2 a x$

Page No 443:

Answer:

Draw AE $⊥$ BC, meeting BC at D.
Applying Pythagoras theorem in right-angled triangle AED, we get:

Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle.
So, BE = CE
and DE+ CE = DE + BE = BD

Page No 443:

Answer:

$We have, ABC as an isosceles triangle, right angled at B . Now, AB = BC$

Applying Pythagoras theorem in right-angled triangle ABC, we get:

${AC}^{2} = {AB}^{2} + {BC}^{2} = 2 {AB}^{2} (∵ AB = AC)$ ...(i)

$∵$ $△ ACD ~ △ ABE$

$We know that ratio of areas of 2 similar triangles is equal to squares of the ratio of their corresponding sides .$

$∴ \frac{ar (△ ABE)}{ar (△ ACD)} = \frac{{AB}^{2}}{{AC}^{2}} = \frac{{AB}^{2}}{2 {AB}^{2}} [from (i)] = \frac{1}{2} = 1 : 2$

Page No 443:

Answer:

Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr
Distance covered by plane A in $1 \frac{1}{2}$ hours = $1000 \times \frac{3}{2} = 1500 km$
Distance covered by plane B in $1 \frac{1}{2}$ hours = $1200 \times \frac{3}{2} = 1800 km$
Now, In right triangle ABC
By using Pythagoras theorem, we have
${AB}^{2} = {BC}^{2} + {CA}^{2} = {(1800)}^{2} + {(1500)}^{2} = 3240000 + 2250000 = 5490000 ∴ {AB}^{2} = 5490000 \Rightarrow AB = 300 \sqrt{61} m$
Hence, the distance between two planes after $1 \frac{1}{2}$ hours is $300 \sqrt{61}$ m.

Page No 443:

Answer:

(a)
In right triangle ALD
Using Pythagoras theorem, we have
AL² = AD² − DL² .....(1)
Again, In right triangle ACL
Using Pythagoras theorem, we have
${AC}^{2} = {AL}^{2} + {LC}^{2} = {AD}^{2} - {DL}^{2} + {(DL + DC)}^{2} [Using (1)] = {AD}^{2} - {DL}^{2} + {(DL + \frac{BC}{2})}^{2} [∵ AD is a median] = {AD}^{2} - {DL}^{2} + {DL}^{2} + {(\frac{BC}{2})}^{2} + BC . DL ∴ {AC}^{2} = {AD}^{2} + BC . DL + {(\frac{BC}{2})}^{2} . . . . . (2)$
(b)
In right triangle ALD
Using Pythagoras theorem, we have
AL² = AD² − DL² .....(3)
Again, In right triangle ABL
Using Pythagoras theorem, we have
${AB}^{2} = {AL}^{2} + {LB}^{2} = {AD}^{2} - {DL}^{2} + {LB}^{2} [Using (3)] = {AD}^{2} - {DL}^{2} + {(BD - DL)}^{2} = {AD}^{2} - {DL}^{2} + {(\frac{1}{2} BC - DL)}^{2} = {AD}^{2} - {DL}^{2} + {(\frac{BC}{2})}^{2} - BC . DL + {DL}^{2} ∴ {AB}^{2} = {AD}^{2} - BC . DL + {(\frac{BC}{2})}^{2} . . . . . (4)$
(c) Adding (2) and (4), we get

${AC}^{2} + {AB}^{2} = {AD}^{2} + BC . DL + {(\frac{BC}{2})}^{2} + {AD}^{2} - BC . DL + {(\frac{BC}{2})}^{2} = 2 {AD}^{2} + \frac{{BC}^{2}}{4} + \frac{{BC}^{2}}{4} = 2 {AD}^{2} + \frac{1}{2} {BC}^{2}$

Page No 443:

Answer:

Naman pulls in the string at the rate of 5 cm per second.
Hence, after 12 seconds the length of the string he will pulled is given by
12 × 5 = 60 cm or 0.6 m
Now, In △BMC
By using Pythagoras theorem, we have
BC² = CM² + MB²
= (2.4)² + (1.8)²
= 9
∴ BC = 3 m
Now, BC' = BC − 0.6
= 3 − 0.6
= 2.4 m
Now, In △BC'M
By using Pythagoras theorem, we have
C'M² = BC'² − MB²
= (2.4)² − (1.8)²
= 2.52
∴ C'M = 1.6 m
The horizontal distance of the fly from him after 12 seconds is given by
C'A = C'M + MA
= 1.6 + 1.2
= 2.8 m

Page No 446:

Answer:

The two triangles are similar if and only if

1. The corresponding sides are in proportion.
2. The corresponding angles are equal.

Page No 446:

Answer:

If a line is drawn parallel to one side of a triangle intersect the other two sides, then it divides the other two sides in the same ratio.

Page No 446:

Answer:

If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Page No 446:

Answer:

The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is equal to one half of the third side.

Page No 446:

Answer:

If the corresponding angles of two triangles are equal, then their corresponding sides are proportional and hence the triangles are similar.

Page No 446:

Answer:

If two angles of a triangle are correspondingly equal to the two angles of another triangle, then the two triangles are similar.

Page No 446:

Answer:

If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.

Page No 446:

Answer:

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.

Page No 446:

Answer:

The square of the hypotenuse is equal to the sum of the squares of the other two sides.
Here, the hypotenuse is the longest side and it's always opposite the right angle.

Page No 446:

Answer:

If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle

Page No 446:

Answer:

By using mid theorem i.e., the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.
$∴ DF ∥ BC and DF = \frac{1}{2} BC \Rightarrow DF = BE$
Since, the opposite sides of the quadrilateral are parallel and equal.
Hence, BDFE is a parallelogram
Similarly, DFCE is a parallelogram.
Now, In △ABC and △EFD
∠ABC = ∠EFD (Opposite angles of a parallelogram)
∠BCA = ∠EDF (Opposite angles of a parallelogram)
By AA similarity criterion, △ABC ∼ △EFD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$∴ \frac{area (△ DEF)}{area (△ ABC)} = {(\frac{DF}{BC})}^{2} = {(\frac{DF}{2 DF})}^{2} = \frac{1}{4}$
Hence, the ratio of the areas of △DEF and △ABC is 1 : 4.

Page No 447:

Answer:

Now, In △ABC and △PQR
∠A = ∠P = 70⁰ (Given)
$\frac{AB}{PQ} = \frac{AC}{PR} [∵ \frac{3}{4.5} = \frac{6}{9} \Rightarrow \frac{1}{1.5} = \frac{1}{1.5}]$
By SAS similarity criterion, △ABC ∼ △PQR

Page No 447:

Answer:

When two triangles are similar, then the ratios of the lengths of their corresponding sides are equal.
Here, △ABC ∼ △DEF
$∴ \frac{AB}{DE} = \frac{BC}{EF} \Rightarrow \frac{AB}{2 AB} = \frac{6}{EF} \Rightarrow EF = 12 cm$

Page No 447:

Answer:

In △ADE and △ABC
∠ADE = ∠ABC (Corresponding angles in DE∥BC)
∠AED = ∠ACB (Corresponding angles in DE∥BC)
By AA similarity criterion, △ADE ∼ △ABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
$∴ \frac{AD}{AB} = \frac{AE}{AC}$
$\Rightarrow \frac{AD}{AD + DB} = \frac{AE}{AE + EC} \Rightarrow \frac{x}{x + 3 x + 4} = \frac{x + 3}{x + 3 + 3 x + 19} \Rightarrow \frac{x}{4 x + 4} = \frac{x + 3}{4 x + 22}$
$\Rightarrow \frac{x}{2 x + 2} = \frac{x + 3}{2 x + 11} \Rightarrow 2 x^{2} + 11 x = 2 x^{2} + 2 x + 6 x + 6 \Rightarrow 3 x = 6 \Rightarrow x = 2$
Hence, the value of x is 2.

Page No 447:

Answer:

Let AB be a ladder and B is the window at 8 m above the ground C.
Now, In right triangle ABC
By using Pythagoras theorem, we have
${AB}^{2} = {BC}^{2} + {CA}^{2} \Rightarrow 10^{2} = 8^{2} + {CA}^{2} \Rightarrow {CA}^{2} = 100 - 64 \Rightarrow {CA}^{2} = 36 \Rightarrow CA = 6 m$
Hence, the distance of the foot of the ladder from the base of the wall is 6 m.

Page No 447:

Answer:

We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides.
Suppose ABC is an equilateral triangle having AB = BC = CA = 2a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC
∴DC = a
Now, In right triangle ADC
By using Pythagoras theorem, we have
${AC}^{2} = {CD}^{2} + {DA}^{2} \Rightarrow {(2 a)}^{2} = a^{2} + {DA}^{2} \Rightarrow {DA}^{2} = 4 a^{2} - a^{2} \Rightarrow {DA}^{2} = 3 a^{2}$
$\Rightarrow DA = \sqrt{3} a$
Hence, the length of the altitude of an equilateral triangle of side 2a cm is $\sqrt{3} a$ cm.

Page No 447:

Answer:

We have △ABC ∼ △DEF
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$∴ \frac{area (△ ABC)}{area (△ DEF)} = {(\frac{BC}{EF})}^{2} \Rightarrow \frac{64}{169} = {(\frac{BC}{EF})}^{2} \Rightarrow {(\frac{8}{13})}^{2} = {(\frac{4}{EF})}^{2}$
$\Rightarrow \frac{8}{13} = \frac{4}{EF} \Rightarrow EF = 6.5 cm$

Page No 447:

Answer:

In △AOB and COD
∠ABO = ∠CDO (Alternte angles in AB∥CD)
∠AOB = ∠COD (Vertically opposite angles)
By AA similarity criterion, △AOB ∼ △COD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$∴ \frac{area (△ AOB)}{area (△ COD)} = {(\frac{AB}{CD})}^{2} \Rightarrow \frac{84}{area (△ COD)} = {(\frac{2 CD}{CD})}^{2} \Rightarrow area (△ COD) = 12 {cm}^{2}$

Page No 447:

Answer:

If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$∴ \frac{area of smaller triangle}{area of larger triangle} = {(\frac{Side of smaller triangle}{Side of larger triangle})}^{2} \Rightarrow \frac{48}{area of larger triangle} = {(\frac{2}{3})}^{2} \Rightarrow area of larger triangle = 108 {cm}^{2}$

Page No 447:

Answer:

We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides..
Suppose ABC is an equilateral triangle having AB = BC = CA = a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC
$∴ DC = \frac{1}{2} a$
Now, In right triangle ADC
By using Pythagoras theorem, we have
${AC}^{2} = {CD}^{2} + {DA}^{2} \Rightarrow a^{2} = {(\frac{1}{2} a)}^{2} + {DA}^{2} \Rightarrow {DA}^{2} = a^{2} - \frac{1}{4} a^{2} \Rightarrow {DA}^{2} = \frac{3}{4} a^{2}$
$\Rightarrow DA = \frac{\sqrt{3}}{2} a Now, area (△ ABC) = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times a \times \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{4} a^{2}$

Page No 447:

Answer:

Suppose ABCD is a rhombus.
We know that the digonals of a rhombus perpendicularly bisect each other.
∴ ∠AOB = 90⁰, AO = 12 cm and BO = 5 cm
Now, In right triangle AOB
By using Pythagoras theorem we have
AB² = AO² + BO²
= 12² + 5²
= 144 + 25
= 169
∴AB² = 169
⇒ AB = 13 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 13 cm

Page No 447:

Answer:

If two traingle are similar then the corresponding angles of the two tringles are equal.
Here, △DEF ∼ △GHK
∴∠E = ∠H = 57^∘
Now, In △DEF
∠D + ∠E + ∠F = 180^∘ (Angle sum property of triangle)
⇒ ∠F = 180^∘ − 48^∘ − 57^∘ = 75^∘

Page No 447:

Answer:

We have
AM : MB = 1 : 2
$\Rightarrow \frac{MB}{AM} = \frac{2}{1}$
Adding 1 to both sides, we get
$\Rightarrow \frac{MB}{AM} + 1 = \frac{2}{1} + 1 \Rightarrow \frac{MB + AM}{AM} = \frac{2 + 1}{1} \Rightarrow \frac{AB}{AM} = \frac{3}{1}$
Now, In △AMN and △ABC
∠AMN = ∠ABC (Corresponding angles in MN∥BC)
∠ANM = ∠ACB (Corresponding angles in MN∥BC)
By AA similarity criterion, △AMN ∼ △ABC
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$∴ \frac{area (△ AMN)}{area (△ ABC)} = {(\frac{AM}{AB})}^{2} = {(\frac{1}{3})}^{2} = \frac{1}{9}$

Page No 447:

Answer:

When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
Here, △BMP ∼ △CNR
$∴ \frac{BM}{CN} = \frac{BP}{CR} = \frac{MP}{NR} . . . . (1) Now, \frac{BM}{CN} = \frac{MP}{NR} [Using (1)] \Rightarrow CN = \frac{BM \times NR}{MP} = \frac{9 \times 9}{6} = 13.5 cm Again, \frac{BM}{CN} = \frac{BP}{CR} [Using (1)] \Rightarrow CR = \frac{BP \times CN}{BM} = \frac{5 \times 13.5}{9} = 7.5 cm$
Perimeter of △CNR = CN + NR + CR = 13.5 + 9 + 7.5 = 30 cm

Page No 447:

Answer:

We know that the altitude drawn from the vertex opposite to the non equal side bisects the non equal side.
Suppose ABC is an isosceles triangle having equal sides AB and BC.
So, the altitude drawn from the vertex will bisect the opposite side.
Now, In right triangle ABD
By using Pythagoras theorem, we have
${AB}^{2} = {BD}^{2} + {DA}^{2} \Rightarrow 25^{2} = 7^{2} + {DA}^{2} \Rightarrow {DA}^{2} = 625 - 49 \Rightarrow {DA}^{2} = 576 \Rightarrow DA = 24 cm$

Page No 448:

Answer:

In right triangle SOW
By using Pythagoras theorem, we have
${OW}^{2} = {WS}^{2} + {SO}^{2} = 35^{2} + 12^{2} = 1225 + 144 = 1369 ∴ {OW}^{2} = 1369 \Rightarrow OW = 37 m$
Hence, the man is 37 m away from the starting point.

Page No 448:

Answer:

Let DC = x
∴ BD = a − x
By using angle bisector theore in △ABC, we have
$\frac{AB}{AC} = \frac{BD}{DC} \Rightarrow \frac{c}{b} = \frac{a - x}{x} \Rightarrow c x = a b - b x \Rightarrow x (b + c) = a b \Rightarrow x = \frac{a b}{(b + c)}$
Now,
$a - x = a - \frac{a b}{b + c} = \frac{a b + a c - a b}{b + c} = \frac{a c}{(a + b)}$

Page No 448:

Answer:

In △AMN and △ABC
∠AMN = ∠ABC = 76⁰ (Given)
∠A = ∠A (Common)
By AA similarity criterion, △AMN ∼ △ABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
$∴ \frac{AM}{AB} = \frac{MN}{BC}$
$\Rightarrow \frac{AM}{AM + MB} = \frac{MN}{BC} \Rightarrow \frac{a}{a + b} = \frac{MN}{c} \Rightarrow MN = \frac{a c}{a + b}$

Page No 448:

Answer:

Suppose ABCD is a rhombus.
We know that the digonals of a rhombus perpendicularly bisect each other.
∴ ∠AOB = 90⁰, AO = 20 cm and BO = 21 cm
Now, In right triangle AOB
By using Pythagoras theorem we have
AB² = AO² + OB²
= 20² + 21²
= 400 + 441
= 841
∴AB² = 841
⇒ AB = 29 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 29 cm

Page No 448:

Answer:

(ii) True
Two circles of any radii are similar to each other.

(i) False
Two rectangles are similar if their corresponding sides are proportional.

(iii) Falase
If two traingles are similar, the their corresponding angles are equal and their corresponding sides are proportional.

(iv) True
Suppose ABC is a triangle and M, N are

Construction: DE is expanded to F sich that EF = DE
To Prove = $DE = \frac{1}{2} BC$
Proof: In △ADE and △CEF
AE = EC        (E is the mid point of AC)
DE = EF        (By construction)
AED = CEF      (Vertically Opposite angle)
By SAS criterion, △ADE ≅ △CEF
CF = AD   (CPCT)
⇒ BD = CF
∠ADE = ∠EFC    (CPCT)
Since, ∠ADE and ∠EFC are alternate angle
Hence, AD ∥ CF and BD ∥ CF
When two sides of a quadrilateral are parallel, then it is a parallelogram
∵DF = BC and BD ∥ CF
∴BDFC is a parallelogram
Hence, DF = BC
⇒ DE + EF = BC
$\Rightarrow DE = \frac{1}{2} BC$

(v) False
In △ABC, AB = 6 cm, ∠A = 45^∘ and AC = 8 cm and in △DEF, DF = 9 cm, ∠D = 45^∘ and DE = 12 cm, then △ABC ∼ △DEF.

In △ABC and △DEF
∠A = ∠D = 45^∘
$\frac{AB}{AC} \neq \frac{DE}{DF}$
So △ABC is not similar to △DEF

(vi) False
The polygon formed by joining the mid points of the sides of a quadrilateral is a parallelogram.

(vii) True

Given: △ABC ∼ △DEF
To Prove = $\frac{Ar (△ ABC)}{Ar (△ DEF)} = {(\frac{AP}{DQ})}^{2}$
Proof: In △ABP and △DEQ
∠BAP = ∠EDQ        (As ∠A = ∠D, so their Half is also equal)
∠B = ∠E         (△ABC ∼ △DEF)
By AA criterion, △ABP ∼ △DEQ
$\frac{AB}{DE} = \frac{AP}{DQ}$            .....(1)
Since, △ABC ∼ △DEF
$∴ \frac{Ar (△ ABC)}{Ar (△ DEF)} = {(\frac{AB}{DE})}^{2} \Rightarrow \frac{Ar (△ ABC)}{Ar (△ DEF)} = {(\frac{AP}{DQ})}^{2} [Using (1)]$

(viii) True

Given: △ABC ∼ △DEF
To Prove = $\frac{Perimeter (△ ABC)}{Perimeter (△ DEF)} = \frac{AP}{DQ}$
Proof: In △ABP and △DEQ
∠B = ∠E          (∵△ABC ∼ △DEF)
∵△ABC ∼ △DEF
$∴ \frac{AB}{DE} = \frac{BC}{EF} \Rightarrow \frac{AB}{DE} = \frac{2 BP}{2 EQ} \Rightarrow \frac{AB}{DE} = \frac{BP}{EQ}$

By SAS criterion, △ABP ∼ △DEQ
$\frac{AB}{DE} = \frac{AP}{DQ}$            .....(1)
Since, △ABC ∼ △DEF
$∴ \frac{Perimeter (△ ABC)}{Perimeter (△ DEF)} = \frac{AB}{DE} \Rightarrow \frac{Perimeter (△ ABC)}{Perimeter (△ DEF)} = \frac{AP}{DQ} [Using (1)]$

(ix) True

Suppose ABCD is a rectangle with O is any point inside it.
Construction: Join OA, OB, OC, OD and draw two parallel lines SQ ∥ AB ∥ DC and PR ∥ BC ∥ AD
To prove: OA² + OC² = OB² + OD²
Proof:
OA² + OC² = (AS² + OS²) + (OQ² + QC²)             [Using Pythagoras theorem in right triangle AOP and COQ]
= (BQ² + OS²) + (OQ² + DS²)
= (BQ² + OQ²) + (OS² + DS²)                                 [Using Pythagoras theorem in right triangle BOQ and DOS]
= OB² + OD²
Hence, LHS = RHS

(x) True

Suppose ABCD is a rhombus having AC and BD its diagonals.
Since, the diagonals of a rhombus perpendicular bisect each other.
Hence, AOC is a right angle triangle
In right triangle AOC
By using Pythagoras theorem, we have
${AB}^{2} = {AO}^{2} + {OB}^{2} \Rightarrow {AB}^{2} = {(\frac{AC}{2})}^{2} + {(\frac{BD}{2})}^{2} [∵ Diagonals of a rhombus perpendicularly bisect each other] \Rightarrow {AB}^{2} = \frac{{AC}^{2}}{4} + \frac{{BD}^{2}}{4} \Rightarrow 4 {AB}^{2} = {AC}^{2} + {BD}^{2} \Rightarrow {AB}^{2} + {AB}^{2} + {AB}^{2} + {AB}^{2} = {AC}^{2} + {BD}^{2} \Rightarrow {AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = {AC}^{2} + {BD}^{2} [∵ All sides of a rhombus are equal]$

Page No 451:

Answer:

(c) 26 m

Suppose, the man starts from point A and goes 24 m due west to point B. From here, he goes 10 m due north and stops at C.
In right triangle ABC, we have:
AB = 24 m, BC = 10 m
Applying Pythagoras theorem, we get:
$A C^{2} = A B^{2} + B C^{2} = 24^{2} + 10^{2} A C^{2} = 576 + 100 = 676 A C = \sqrt{676} = 26$

Page No 451:

Answer:

(b) 10 m

Let AB and DE be the two poles.
According to the question:
AB = 13 m
DE = 7 m
Distance between their bottoms = BE = 8 m
Draw a perpendicular DC to AB from D, meeting AB at C. We get:
DC = 8m, AC = 6 m
Applying Pythagoras theorem in right-angled triangle ACD, we have:
$A D^{2} = D C^{2} + A C^{2} = 8^{2} + 6^{2} = 64 + 36 = 100 A D = \sqrt{100} = 10 m$

Page No 451:

Answer:

(c) 1.5 m

Let AB and AC be the vertical stick and its shadow, respectively.
According to the question:
AB = 1.8 m
AC = 45 cm = 0.45 m
Again, let DE and DF be the pole and its shadow, respectively.
According to the question:
DE = 6 m
DF = ?
Now, in right-angled triangles ABC and DEF, we have:
$∠ B A C = ∠ E D F = 90 ° ∠ A C B = ∠ D F E (Angular elevation of the Sun at the same time) Therefore, by AA similarity theorem, we get : △ A B C ~ △ D E F$

$\Rightarrow \frac{A B}{A C} = \frac{D E}{D F} \Rightarrow \frac{1.8}{0.45} = \frac{6}{D F} \Rightarrow D F = \frac{6 \times 0.45}{1.8} = 1.5 m$

Page No 451:

Answer:

(d) 30 m

Let AB and AC be the vertical pole and its shadow, respectively.
According to the question:
AB = 6 m
AC = 3.6 m
Again, let DE and DF be the tower and its shadow.
According to the question:
DF = 18 m
DE = ?
Now, in right-angled triangles ABC and DEF, we have:
$∠ B A C = ∠ E D F = 90 ° ∠ A C B = ∠ D F E (Angular elevation of the Sun at the same time) Therefore, by AA similarity theorem, we get : △ A B C ~ △ D E F$

$\Rightarrow \frac{A B}{A C} = \frac{D E}{D F} \Rightarrow \frac{6}{3.6} = \frac{D E}{18} \Rightarrow D E = \frac{6 \times 18}{3.6} = 30 m$

Page No 451:

Answer:

Suppose DE is a 5 m long stick and BC is a 12.5 m high tree.
Suppsose DA and BA are the shadows of DE and BC respectively.
Now, In △ABC and △ADE
∠ABC= ∠ADE = 90⁰
∠A = ∠A (Common)
By AA-similarity criterion
△ABC ∼ △ADE
If two triangles are similar, then the the ratio of their corresponding sides are equal.
$∴ \frac{AB}{AD} = \frac{BC}{DE} \Rightarrow \frac{AB}{2} = \frac{12.5}{5} \Rightarrow AB = 5 cm$
Hence, the correct answer is option (d).

Page No 451:

Answer:

(a) 7 m

Let the ladder BC reaches the building at C.
Let the height of building where the ladder reaches be AC.
According to the question:
BC = 25 m
AC = 24 m
In right-angled triangle CAB, we apply Pythagoras theorem to find the value of AB.
$B C^{2} = A C^{2} + A B^{2} \Rightarrow A B^{2} = B C^{2} - A C^{2} = 25^{2} - 24^{2} \Rightarrow A B^{2} = 625 - 576 = 49 \Rightarrow A B = \sqrt{49} = 7 m$

Page No 451:

Answer:

Now, In right triangle MOP
By using Pythagoras theorem, we have
${MP}^{2} = {PO}^{2} + {OM}^{2} = 12^{2} + 16^{2} = 144 + 256 = 400 ∴ {MP}^{2} = 400 \Rightarrow MP = 20 cm$
Now, In right triangle MPN
By using Pythagoras theorem, we have
${PN}^{2} = {NM}^{2} + {MP}^{2} = 21^{2} + 20^{2} = 441 + 400 = 841 ∴ {MP}^{2} = 841 \Rightarrow MP = 29 cm$
Hence, the correct answer is option (b).

Page No 451:

Answer:

(b) 15 cm, 20 cm

It is given that the length of hypotenuse is 25 cm.
Let the other two sides be x cm and (x $-$ 5) cm.
Applying Pythagoras theorem, we get:

$25^{2} = x^{2} + (x - 5)^{2} \Rightarrow 625 = x^{2} + x^{2} + 25 - 10 x \Rightarrow 2 x^{2} - 10 x - 600 = 0 \Rightarrow x^{2} - 5 x - 300 = 0 \Rightarrow x^{2} - 20 x + 15 x - 300 = 0 \Rightarrow x (x - 20) + 15 (x - 20) = 0 \Rightarrow (x - 20) (x + 15) = 0 \Rightarrow x - 20 = 0 o r x + 15 = 0 \Rightarrow x = 20 or x = - 15 Side of a triangle cannot be negative . Therefore, x = 20 cm$

Now,
x $-$ 5 = 20 $-$ 5 = 15 cm

Page No 451:

Answer:

(b) $6 \sqrt{3} cm$

Let ABC be the equilateral triangle with AD as its altitude from A.

In right-angled triangle ABD, we have:

$A B^{2} = A D^{2} + B D^{2} A D^{2} = A B^{2} - B D^{2} = 12^{2} - 6^{2} = 144 - 36 = 108 A D = \sqrt{108} = 6 \sqrt{3} cm$

Page No 452:

Answer:

(d) 24 cm

In triangle ABC, let the altitude from A on BC meets BC at D.
We have:
AD = 5 cm, AB = 13 cm and D is the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:
$A B^{2} = A D^{2} + B D^{2} \Rightarrow B D^{2} = A B^{2} - A D^{2} \Rightarrow B D^{2} = 13^{2} - 5^{2} \Rightarrow B D^{2} = 169 - 25 \Rightarrow B D^{2} = 144 \Rightarrow B D = \sqrt{144} = 12 cm$

Therefore, BC = 2BD = 24 cm

Page No 452:

Answer:

(a) 3 : 4

In ∆ ABD and ∆ACD, we have:
$∠ B A D = ∠ C A D$

Now,

$\frac{B D}{D C} = \frac{A B}{A C} = \frac{6}{8} = \frac{3}{4} B D : D C = 3 : 4$

Page No 452:

Answer:

(d) 7.5 cm
It is given that AD bisects angle A.
Therefore, applying angle bisector theorem, we get:

$\frac{B D}{D C} = \frac{A B}{A C} \Rightarrow \frac{4}{5} = \frac{6}{x} \Rightarrow x = \frac{5 \times 6}{4} = 7.5$

Hence, AC = 7.5 cm

Page No 452:

Answer:

By using angle bisector theore in △ABC, we have
$\frac{AB}{AC} = \frac{BD}{DC} \Rightarrow \frac{10}{14} = \frac{6 - x}{x} \Rightarrow 10 x = 84 - 14 x \Rightarrow 24 x = 84 \Rightarrow x = 3.5$
Hence, the correct answer is option (b).

Page No 452:

Answer:

(b) isosceles

In an isosceles triangle, the perpendicular from the vertex to the base bisects the base.

Page No 452:

Answer:

(c) 3AB² = 4AD²

Applying Pythagoras theorem in right-angled triangles ABD and ADC, we get:

$A B^{2} = A D^{2} + B D^{2} \Rightarrow A B^{2} = {(\frac{1}{2} A B)}^{2} + A D^{2} (∵ △ ABC is equilateral and A D = \frac{1}{2} AB) \Rightarrow A B^{2} = \frac{1}{4} A B^{2} + A D^{2} \Rightarrow A B^{2} - \frac{1}{4} A B^{2} = A D^{2} \Rightarrow \frac{3}{4} A B^{2} = A D^{2} \Rightarrow 3 A B^{2} = 4 A D^{2}$

Page No 452:

Answer:

(c) 16 cm

Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
Also, diagonals of a rhombus bisect each other at right angles.
If AC =12 cm, AO = 6 cm
Applying Pythagoras theorem in right-angled triangle AOB. we get:

$A B^{2} = A O^{2} + B O^{2} \Rightarrow B O^{2} = A B^{2} - A O^{2} \Rightarrow B O^{2} = 10^{2} - 6^{2} = 100 - 36 = 64 \Rightarrow B O = \sqrt{64} = 8 \Rightarrow B D = 2 \times B O = 2 \times 8 = 16 cm$

Hence, the length of the second diagonal BD is 16 cm.

Page No 452:

Answer:

(b) 13 cm

Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We have:
AC = 24 cm and BD = 10 cm
We know that diagonals of a rhombus bisect each other at right angles.
Therefore applying Pythagoras theorem in right-angled triangle AOB, we get:
$A B^{2} = A O^{2} + B O^{2} = 12^{2} + 5^{2} = 144 + 25 = 169 A B = \sqrt{169} = 13$

Hence, the length of each side of the rhombus is 13 cm.

Page No 453:

Answer:

(b) trapezium
Diagonals of a trapezium divide each other proportionally.

Page No 453:

Answer:

(a) 2

We know that the diagonals of a trapezium are proportional.
Therefore,

$\frac{O A}{O C} = \frac{O B}{O D} \Rightarrow \frac{3 X - 1}{5 X - 3} = \frac{2 X + 1}{6 X - 5} \Rightarrow (3 X - 1) (6 X - 5) = (2 X + 1) (5 X - 3) \Rightarrow 18 X^{2} - 15 X - 6 X + 5 = 10 X^{2} - 6 X + 5 X - 3 \Rightarrow 18 X^{2} - 21 X + 5 = 10 X^{2} - X - 3 \Rightarrow 18 X^{2} - 21 X + 5 - 10 X^{2} + X + 3 = 0 \Rightarrow 8 X^{2} - 20 X + 8 = 0 \Rightarrow 4 (2 X^{2} - 5 X + 2) = 0 \Rightarrow 2 X^{2} - 5 X + 2 = 0 \Rightarrow 2 X^{2} - 4 X - X + 2 = 0 \Rightarrow 2 X (X - 2) - 1 (X - 2) = 0 \Rightarrow (X - 2) (2 X - 1) = 0 \Rightarrow Either x - 2 = 0 or 2 x - 1 = 0 \Rightarrow Either x = 2 or x = \frac{1}{2} When x = \frac{1}{2}, 6 x - 5 = - 2 < 0, which is not possible . Therefore, x = 2$

Page No 453:

Answer:

(a) a parallelogram

The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.

Page No 453:

Answer:

(c) isosceles

Let AD be the angle bisector of angle A in triangle ABC.
Applying angle bisector theorem, we get:
$\frac{A B}{A C} = \frac{B D}{D C}$

It is given that AD bisects BC.
Therefore, BD = DC
$\Rightarrow \frac{A B}{A C} = 1 \Rightarrow A B = A C$

Therefore, the triangle is isosceles.

Page No 453:

Answer:

(a) 30 $°$
We have:

$\frac{A B}{A C} = \frac{B D}{D C}$

Applying angle bisector theorem, we can conclude that AD bisects $∠$ A.

$In △ A B C, ∠ A + ∠ B + ∠ C = 180 ° \Rightarrow ∠ A = 180 - ∠ B - ∠ C \Rightarrow ∠ A = 180 - 70 - 50 = 60 ° ∵ ∠ B A D = ∠ C A D = \frac{1}{2} ∠ B A C ∴ ∠ B A D = \frac{1}{2} \times 60 = 30 °$

Page No 453:

Answer:

(b) 6 cm

It is given that DE $∥$ BC.

Applying basic proportionality theorem, we have:

$\frac{A D}{B D} = \frac{A E}{E C} \Rightarrow \frac{2.4}{B D} = \frac{3.2}{4.8} \Rightarrow B D = \frac{2.4 \times 4.8}{3.2} = 3.6 cm$

Therefore, AB = AD + BD = 2.4 + 3.6 = 6 cm

Page No 453:

Answer:

(b) 4 cm

It is given that DE $∥$ BC.

Applying basic proportionality theorem, we get:

$\frac{A D}{A B} = \frac{A E}{A C} \Rightarrow \frac{4.5}{7.2} = \frac{A E}{6.4} \Rightarrow A E = \frac{4.5 \times 6.4}{7.2} = 4 cm$

Page No 454:

Answer:

(c) x = 4

It is given that $D E ∥ B C$ .
Applying Thales' theorem, we get:

$\frac{A D}{B D} = \frac{A E}{E C} \Rightarrow \frac{7 x - 4}{3 x + 4} = \frac{5 x - 2}{3 x} \Rightarrow 3 x (7 x - 4) = (5 x - 2) (3 x + 4) \Rightarrow 21 x^{2} - 12 x = 15 x^{2} + 20 x - 6 x - 8 \Rightarrow 21 x^{2} - 12 x = 15 x^{2} + 14 x - 8 \Rightarrow 21 x^{2} - 12 x - 15 x^{2} - 14 x + 8 = 0 \Rightarrow 6 x^{2} - 26 x + 8 = 0 \Rightarrow 2 (3 x^{2} - 13 x + 4) = 0 \Rightarrow 3 x^{2} - 13 x + 4 = 0 \Rightarrow 3 x^{2} - 12 x - x + 4 = 0 \Rightarrow 3 x (x - 4) - 1 (x - 4) = 0 \Rightarrow (x - 4) (3 x - 1) = 0 \Rightarrow x - 4 = 0 or 3 x - 1 = 0 \Rightarrow x = 4 or x = \frac{1}{3} If x = \frac{1}{3}, 7 x - 4 = - \frac{5}{3} < 0; it is not possible . Therefore, x = 4$

Page No 454:

Answer:

(d) 2.1 cm

It is given that DE $∥ B C$ .
Applying Thales' theorem, we get:
$\frac{A D}{D B} = \frac{A E}{E C} Let A E be x cm . Therefore, E C = (5.6 - x) cm \Rightarrow \frac{3}{5} = \frac{x}{5.6 - x} \Rightarrow 3 (5.6 - x) = 5 x \Rightarrow 16.8 - 3 x = 5 x \Rightarrow 8 x = 16.8 \Rightarrow x = 2.1 cm$

Page No 454:

Answer:

(b) 5.4 cm

∆ABC ∼ ∆DEF

Therefore,

$\frac{Perimeter (△ A B C)}{Perimeter (△ D E F)} = \frac{B C}{E F} \Rightarrow \frac{30}{18} = \frac{9}{E F} \Rightarrow E F = \frac{9 \times 18}{30} = 5.4 cm$

Page No 454:

Answer:

(a) 35 cm

$∵$ ∆ABC ∼ ∆DEF

$∴ \frac{Perimeter (△ A B C)}{Perimeter (△ D E F)} = \frac{A B}{D E} \Rightarrow \frac{Perimeter (△ A B C)}{25} = \frac{9.1}{6.5} \Rightarrow P erimeter (△ A B C) = \frac{9.1 \times 25}{6.5} = 35 cm$

Page No 454:

Answer:

(d) 30 cm

Perimeter of $△$ ABC = AB + BC + CA = 9 + 6 + 7.5 = 22.5 cm

$∵$ ∆DEF ∼ ∆ABC

$∴ \frac{Perimeter △ (A B C)}{Perimeter (△ D E F)} = \frac{B C}{E F} \Rightarrow \frac{22.5}{Perimeter (△ D E F)} = \frac{6}{8} \Rightarrow Perimeter (△ D E F) = \frac{22.5 \times 8}{6} = 30 cm$

Page No 454:

Answer:

Given: ABC and BDE are two equilateral triangles
Since, D is the mid point of BC and BDE is also an equilateral traingle.
Hence, E is also the mid point of AB.
Now, D and E are the mid points of BC and AB.
In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it.
$DE ∥ CA and DE = \frac{1}{2} CA$
Now, In △ABC and △EBD
∠BED = ∠BAC (Corresponding angles)
∠B = ∠B (Common)
By AA-similarity criterion
△ABC ∼ △EBD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$∴ \frac{area (△ ABC)}{area (△ DBE)} = {(\frac{AC}{ED})}^{2} = {(\frac{2 ED}{ED})}^{2} = \frac{4}{1}$
Hence, the correct answer is option (d).

Page No 454:

Answer:

(b) DE = 12 cm, $∠$ F = 100°

Disclaimer: In the question, it should be ∆ABC ∼ ∆DFE instead of ∆ABC ∼ ∆DEF.

In triangle ABC,
$∠ A + ∠ B + ∠ C = 180 ° ∴ ∠ B = 180 - 30 - 50 = 100 °$

$∵$ ∆ABC ∼ ∆DFE

$∴ ∠ D = ∠ A = 30 °, ∠ F = ∠ B = 100 ° and ∠ E = ∠ C = 50 ° Also, \frac{A B}{D F} = \frac{A C}{D E} \Rightarrow \frac{5}{7.5} = \frac{8}{D E} \Rightarrow D E = \frac{8 \times 7.5}{5} = 12 cm$

Page No 454:

Answer:

(c) BD ⋅ CD = AD²

$In △ B D A and △ A D C, we have : ∠ B D A = ∠ A D C = 90 ° ∠ A B D = 90 ° - ∠ D A B = 90 ° - (90 ° - ∠ D A C) = 90 ° - 90 ° + ∠ D A C = ∠ D A C Applying A A similarity theorem, we conclude that △ B D A ~ △ A D C . \Rightarrow \frac{B D}{A D} = \frac{A D}{C D} \Rightarrow A D^{2} = B D . C D$

Page No 454:

Answer:

$AB = 6 \sqrt{3} cm \Rightarrow {AB}^{2} = 108 {cm}^{2} AC = 12 cm \Rightarrow {AC}^{2} = 144 {cm}^{2} BC = 6 cm \Rightarrow {BC}^{2} = 36 cm ∴ {AC}^{2} = {AB}^{2} + {BC}^{2}$
Since, the square of the longest side is equal to the sum of the square of two sides, so △ABC is a right angled triangle.
∴The angle opposite to AC i.e. ∠B = 90^o
Hence, the correct answer is option (c)

Page No 455:

Answer:

(c) $∠$ B = $∠$ D

$Disclaimer : In the question, the ratio should be \frac{A B}{D E} = \frac{B C}{F D} = \frac{A C}{E F} . We can write it as : \frac{A B}{E D} = \frac{B C}{D F} = \frac{A C}{F E} Therefore, △ A B C ~ E D F Hence, the corresponding angles, i . e ., ∠ B and ∠ D, will be equal . i . e ., ∠ B = ∠ D$

Page No 455:

Answer:

(b) $\frac{D E}{P Q} = \frac{E F}{R P}$

In ∆DEF and ∆PQR, we have:

$∠ D = ∠ Q and ∠ R = ∠ E Applying A A similarity theorem, we conclude that △ D E F ~ △ Q R P . Hence, \frac{D E}{Q R} = \frac{D F}{Q P} = \frac{E F}{P R}$

Page No 455:

Answer:

(c) BC ⋅ DE = AB ⋅ EF

∆ABC ∼ ∆EDF
Therefore,
$\frac{A B}{D E} = \frac{A C}{E F} = \frac{B C}{D F} \Rightarrow B C . D E \neq A B . E F$

Page No 455:

Answer:

(b) similar but not congruent

In ∆ABC and ∆DEF, we have:
$∠ B = ∠ E and ∠ F = ∠ C Applying A A similarity theorem, we conclude that △ A B C ~ △ D E F . Also, A B = 3 D E \Rightarrow A B \neq D E Therefore, △ A B C and △ D E F are not congruent .$

Page No 455:

Answer:

(a) ∆PQR ∼ ∆CAB

In ∆ABC and ∆PQR, we have:
$\frac{A B}{Q R} = \frac{B C}{P R} = \frac{C A}{P Q} \Rightarrow △ A B C ~ △ Q R P We can also write it as △ P Q R ~ △ C A B .$

Page No 455:

Answer:

(d) 100°

$In △ A P B and △ D P C, we have : ∠ A P B = ∠ D P C = 50 ° \frac{A P}{B P} = \frac{6}{3} = 2 \frac{D P}{C P} = \frac{5}{2.5} = 2 Hence, \frac{A P}{B P} = \frac{D P}{C P} Applying S A S theorem, we conclude that △ A P B ~ △ D P C . ∴ ∠ P B A = ∠ P C D In △ D P C, we have : ∠ C D P + ∠ C P D + ∠ P C D = 180 ° \Rightarrow ∠ P C D = 180 ° - ∠ C D P - ∠ C P D \Rightarrow ∠ P C D = 180 ° - 30 ° - 50 ° \Rightarrow ∠ P C D = 100 ° Therefore, ∠ P B A = 100 °$

Page No 455:

Answer:

If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$∴ \frac{area of first triangle}{area of second triangle} = {(\frac{Side of first triangle}{Side of second triangle})}^{2} = {(\frac{4}{9})}^{2} = \frac{16}{81}$
Hence, the correct answer is option (d).

Page No 455:

Answer:

(d) 9 : 4
It is given that ∆ ABC ∼ ∆PQR and $\frac{B C}{Q R} = \frac{2}{3}$ .
Therefore,
$\frac{a r (△ P Q R)}{a r (△ A B C)} = \frac{Q R^{2}}{B C^{2}} = {(\frac{3}{2})}^{2} = \frac{9}{4}$

Page No 455:

Answer:

(b) 4:1
In ∆ABC, D is the midpoint of AB and E is the midpoint of AC.
Therefore, by midpoint theorem, DE $∥$ BC.
Also, by Basic Proportionality Theorem,
$\frac{A D}{D B} = \frac{A E}{E C}$

$A l s o, A B = A C = B C (∵ ∆ A B C is an equilateral triangle) S o, \frac{A D}{D B} = \frac{A E}{E C} = 1 In △ A B C a n d △ A D E, we have : ∠ A = ∠ A \frac{A D}{A B} = \frac{A E}{A C} = \frac{1}{2} ∴ △ A B C ~ △ A D E (S A S criterion) ∴ a r (△ A B C) : a r (△ A D E) = {(A B)}^{2} : {(A D)}^{2} \Rightarrow a r (△ A B C) : a r (△ A D E) = 2^{2} : 1^{2} \Rightarrow a r (△ A B C) : a r (△ A D E) = 4 : 1$

Page No 456:

Answer:

(b) 25 : 49

$In △ ABC and △ DEF, we have : \frac{A B}{D E} = \frac{B C}{E F} = \frac{A C}{D F} = \frac{5}{7} Therefore, by SSS criterion, we conclude that △ A B C ~ △ D E F . \Rightarrow \frac{a r (△ A B C)}{a r (△ D E F)} = \frac{A B^{2}}{D E^{2}} = {(\frac{5}{7})}^{2} = \frac{25}{49} = 25 : 49$

Page No 456:

Answer:

(b) 6:7

$∵$ ∆ABC ∼ ∆DEF

$∴ \frac{A B}{D E} = \frac{B C}{E F} = \frac{A C}{D F} . . . (i) Also, \frac{a r (△ A B C)}{a r (△ D E F)} = \frac{A B^{2}}{D E^{2}} \Rightarrow \frac{36}{49} = \frac{A B^{2}}{D E^{2}} \Rightarrow \frac{6}{7} = \frac{A B}{D E} \Rightarrow \frac{A B}{D E} = \frac{B C}{E F} = \frac{A C}{D F} = \frac{6}{7} (from (i)) Thus, the ratio of corresponding sides is 6 : 7 .$

Page No 456:

Answer:

(c) 5:6
Let x and y be the corresponding heights of the two triangles.
It is given that the corresponding angles of the triangles are equal.
Therefore, the triangles are similar. (By AA criterion)
Hence,
$\frac{a r (△_{1})}{a r (△_{2})} = \frac{25}{36} = \frac{x^{2}}{y^{2}} \Rightarrow \frac{x^{2}}{y^{2}} = \frac{25}{36} \Rightarrow \frac{x}{y} = \sqrt{\frac{25}{36}} = \frac{5}{6} = 5 : 6$

Page No 456:

Answer:

(b) similar to the original triangle

The line segments joining the midpoints of the sides of a triangle form four triangles, each of which is similar to the original triangle.

Page No 456:

Answer:

(b) 10 cm

$∵ △ A B C ~ △ Q R P ∴ \frac{A B}{Q R} = \frac{B C}{P R} Now, \frac{a r (△ A B C)}{a r (△ Q R P)} = \frac{9}{4} \Rightarrow (\frac{A B}{Q R})^{2} = \frac{9}{4} \Rightarrow \frac{A B}{Q R} = \frac{3}{2} Therefore, \frac{A B}{Q R} = \frac{B C}{P R} = \frac{3}{2} Hence, 3 P R = 2 B C = 2 \times 15 = 30 P R = 10 cm$

Page No 456:

Answer:

(c) isosceles and similar

In ∆AOC and ∆ODB, we have:

$∠ A O C = ∠ D O B (Vertically opposite angles) a n d ∠ O A C = ∠ O D B (Angles in the same segment) Therefore, by AA similarity theorem, we conclude that △ A O C ~ △ D O B . \Rightarrow \frac{O C}{O B} = \frac{O A}{O D} = \frac{A C}{B D} Now, O B = O D \Rightarrow \frac{O C}{O A} = \frac{O B}{O D} = 1 \Rightarrow O C = O A Hence, △ OAC and △ ODB are isosceles and similar .$

Page No 456:

Answer:

(d) 90°

Given:
AC = BC
$A B^{2} = 2 A C^{2} = A C^{2} + A C^{2} = A C^{2} + B C^{2}$
Applying Pythagoras theorem, we conclude that ∆ABC is right angled at C.
or, $∠ C = 90 °$

Page No 456:

Answer:

(b) right-angled

We have:
$A B^{2} + B C^{2} = 16^{2} + 12^{2} = 256 + 144 = 400 and, A C^{2} = 20^{2} = 400 ∴ A B^{2} + B C^{2} = A C^{2}$

Hence, ∆ABC is a right-angled triangle.

Page No 456:

Answer:

(c) Two triangles are similar if their corresponding sides are proportional.
According to the statement:
$△ A B C ~ △ D E F if \frac{A B}{D E} = \frac{A C}{D F} = \frac{B C}{E F}$

Page No 457:

Answer:

(b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding sides.

Because the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Page No 457:

Answer:

(a) - (s)
Let AE be x.
Therefore, EC = 5.6 $-$ x
It is given that DE $∥$ BC.
Therefore, by B.P.T., we get:

$\frac{A D}{D B} = \frac{A E}{E C} \Rightarrow \frac{3}{5} = \frac{x}{5.6 - x} \Rightarrow 3 (5.6 - x) = 5 x \Rightarrow 16.8 - 3 x = 5 x \Rightarrow 8 x = 16.8 \Rightarrow x = 2.1 cm$

(b) - (q)

$∵ △ A B C ~ △ D E F ∴ \frac{A B}{D E} = \frac{B C}{E F} \Rightarrow \frac{3}{2} = \frac{6}{E F} E F = \frac{6 \times 2}{3} = 4 cm$

(c) - (p)

$∵ △ A B C ~ △ P Q R ∴ \frac{a r (△ A B C)}{a r (△ P Q R)} = \frac{B C^{2}}{Q R^{2}} \Rightarrow \frac{9}{16} = \frac{4 . 5^{2}}{Q R^{2}} \Rightarrow Q R = \sqrt{\frac{4.5 \times 4.5 \times 16}{9}} = \frac{4.5 \times 4}{3} = 6 cm$

(d) - (r)

$∵ A B ∥ C D ∴ \frac{O A}{O B} = \frac{O C}{O D} (Thales' theorem) \Rightarrow \frac{2 x + 4}{9 x - 21} = \frac{2 x - 1}{3} \Rightarrow 3 (2 x + 4) = (2 x - 1) (9 x - 21) \Rightarrow 6 x + 12 = 18 x^{2} - 42 x - 9 x + 21 \Rightarrow 18 x^{2} - 57 x + 9 = 0 \Rightarrow 6 x^{2} - 19 x + 3 = 0 \Rightarrow 6 x^{2} - 18 x - x + 3 = 0 \Rightarrow (6 x - 1) (x - 3) = 0 \Rightarrow x = 3 o r x = - \frac{1}{6} But x = - \frac{1}{6} makes (2 x - 1) < 0, which is not possible . Therefore, x = 3$

Page No 458:

Answer:

(a) - (r)
Let the man starts from A and goes 10 m due east at B and then 20 m due north at C.
Then, in right-angled triangle ABC, we have:
$A B^{2} + B C^{2} = A C^{2} \Rightarrow A C = \sqrt{10^{2} + 20^{2}} = \sqrt{100 + 200} = 10 \sqrt{3}$
Hence, the man is 10 $\sqrt{3}$ m away from the starting point.

(b) - (q)
Let the triangle be ABC with altitude AD.
In right-angled triangle ABD, we have:
$A B^{2} = A D^{2} + B D^{2} \Rightarrow A D^{2} = 10^{2} - 5^{2} (∵ B D = \frac{1}{2} B C) \Rightarrow A D = \sqrt{100 - 25} = \sqrt{75} = 5 \sqrt{3} cm$

(c) - (p)
Area of an equilateral triangle with side a = $\frac{\sqrt{3}}{4} a^{2} = \frac{\sqrt{3}}{4} \times 10^{2} = \sqrt{3} \times 5 \times 5 = 25 \sqrt{3} {cm}^{2}$

(d) - (s)
Let the rectangle be ABCD with diagonals AC and BD.
In right-angled triangle ABC, we have:
$A C^{2} = A B^{2} + B C^{2} = 8^{2} + 6^{2} = 64 + 36 \Rightarrow A C = \sqrt{100} = 10 m$

Page No 462:

Answer:

(b) 7.5 cm

$∵$ ∆ABC ∼ ∆DEF

$∴ \frac{Perimeter (△ ABC)}{Perimeter (△ DEF)} = \frac{AB}{DE} \Rightarrow \frac{32}{24} = \frac{10}{D E}$

$\Rightarrow D E = \frac{10 \times 24}{32} = 7.5 cm$

Page No 462:

Answer:

(a) 5.6 cm
$∵$ DE ∥ BC

$∴ \frac{A D}{A B} = \frac{A E}{A C} = \frac{D E}{B C} (Thales' theorem) \Rightarrow \frac{3.5}{A B} = \frac{5}{8} \Rightarrow A B = \frac{3.5 \times 8}{5} = 5.6 cm$

Page No 462:

Answer:

(b) 13 m

Let the poles be AB and CD.
It is given that:
AB = 6 m and CD = 11 m
Let AC be 12 m.
Draw a perpendicular from B on CD, meeting CD at E.
Then,
BE = 12 m
We have to find BD.
Applying Pythagoras theorem in right-angled triangle BED, we have:
$B D^{2} = B E^{2} + E D^{2} = 12^{2} + 5^{2} (∵ E D = C D - C E = 11 - 6) = 144 + 25 = 169 B D = 13 m$

Page No 462:

Answer:

(c)
We know that the ratio of areas of similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let h be the altitude of the other triangle.
Therefore,

$\frac{25}{36} = \frac{{(3.5)}^{2}}{h^{2}}$ ^{$\Rightarrow h^{2} = \frac{{(3.5)}^{2} \times 36}{25} \Rightarrow h^{2} = 17.64 \Rightarrow h = 4.2 cm$}

Page No 463:

Answer:

$∵$ ∆ABC ∼ ∆DEF

$∴ \frac{A B}{D E} = \frac{B C}{E F} \Rightarrow \frac{1}{2} = \frac{6}{E F} \Rightarrow E F = 12 cm$

Page No 463:

Answer:

$∵$ DE ∥ BC

$∴ \frac{A D}{D B} = \frac{A E}{E C} (Basic proportionality theorem) \frac{x}{3 x + 4} = \frac{x + 3}{3 x + 19} \Rightarrow x (3 x + 19) = (x + 3) (3 x + 4) \Rightarrow 3 x^{2} + 19 x = 3 x^{2} + 4 x + 9 x + 12 \Rightarrow 19 x - 13 x = 12 \Rightarrow 6 x = 12 \Rightarrow x = 2$

Page No 463:

Answer:

Let the ladder be AB and BC be the height of the window from the ground.

We have:
AB = 10 m and BC = 8 m
Applying Pythagoras theorem in right-angled triangle ACB, we have:

$A B^{2} = A C^{2} + B C^{2} \Rightarrow A C^{2} = A B^{2} - B C^{2} = 10^{2} - 8^{2} = 100 - 64 = 36 \Rightarrow A C = 6 m$

Hence, the foot of the ladder is 6 m away from the base of the wall.

Page No 463:

Answer:

Let the triangle be ABC with AD as its altitude. Then, D is the midpoint of BC.
In right-angled triangle ABD, we have:

$A B^{2} = A D^{2} + D B^{2} \Rightarrow A D^{2} = A B^{2} - D B^{2} = 4 a^{2} - a^{2} (∵ B D = \frac{1}{2} B C) = 3 a^{2} A D = \sqrt{3} a$

Hence, the length of the altitude of an equilateral triangle of side 2a cm is $\sqrt{3} a$ cm.

Page No 463:

Answer:

$∵$ ∆ABC ∼ ∆DEF

$∴ \frac{a r (△ A B C)}{a r (△ D E F)} = \frac{B C^{2}}{E F^{2}} \Rightarrow \frac{64}{169} = \frac{4^{2}}{E F^{2}} \Rightarrow E F^{2} = \frac{16 \times 169}{64} \Rightarrow E F = \frac{4 \times 13}{8} = 6.5 cm$

Page No 463:

Answer:

In ∆AOB and ∆COD, we have:

$∠ A O B = ∠ C O D (Vertically opposite angles) ∠ OAB = ∠ OCD (Alternate angles as AB ∥ CD) Applying AA similiarity criterion, we get : △ AOB ~ △ COD ∴ \frac{a r (△ A O B)}{a r (△ C O D)} = \frac{A B^{2}}{C D^{2}} \Rightarrow \frac{84}{a r (△ C O D)} = {(\frac{A B}{C D})}^{2} \Rightarrow \frac{84}{a r (△ C O D)} = {(\frac{2 C D}{C D})}^{2} \Rightarrow a r (△ C O D) = \frac{84}{4} = 21 {cm}^{2}$

Page No 463:

Answer:

It is given that the triangles are similar.
Therefore, the ratio of areas of similar triangles will be equal to the ratio of squares of their corresponding sides.

$∴ \frac{48}{Area of larger triangle} = \frac{2^{2}}{3^{2}} \Rightarrow \frac{48}{Area of larger triangle} = \frac{4}{9} \Rightarrow Area of larger triangle = \frac{48 \times 9}{4} = 108 {cm}^{2}$

Page No 463:

Answer:

$L M ∥ C B and L N ∥ C D$

Therefore, applying Thales' theorem, we have:
$\frac{A B}{A M} = \frac{A C}{A L} and \frac{A D}{A N} = \frac{A C}{A L} \Rightarrow \frac{A B}{A M} = \frac{A D}{A N} ∴ \frac{A M}{A B} = \frac{A N}{A D}$
This completes the proof.

Page No 463:

Answer:

Let the triangle be ABC with AD as the bisector of $∠ A$ which meets BC at D.
We have to prove:
$\frac{B D}{D C} = \frac{A B}{A C}$

Draw CE $∥$ DA, meeting BA produced at E.
CE $∥$ DA
Therefore,
$∠ 2 = ∠ 3 (Alternate angles) and ∠ 1 = ∠ 4 (Corresponding angles) But, ∠ 1 = ∠ 2 Therefore, ∠ 3 = ∠ 4 \Rightarrow A E = A C In △ B C E, D A ∥ C E . Applying Thales' theorem, we gave : \frac{B D}{D C} = \frac{A B}{A E} \Rightarrow \frac{B D}{D C} = \frac{A B}{A C}$
This completes the proof.

Page No 463:

Answer:

Let ABC be the equilateral triangle with each side equal to a.
Let AD be the altitude from A, meeting BC at D.
Therefore, D is the midpoint of BC.
Let AD be h.
Applying Pythagoras theorem in right-angled triangle ABD, we have:
$A B^{2} = A D^{2} + B D^{2} \Rightarrow a^{2} = h^{2} + (\frac{a}{2})^{2} \Rightarrow h^{2} = a^{2} - \frac{a^{2}}{4} = \frac{3}{4} a^{2} \Rightarrow h = \frac{\sqrt{3}}{2} a$

Therefore,
$Area of triangle ABC = \frac{1}{2} \times base \times height = \frac{1}{2} \times a \times \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{4} a^{2}$

This completes the proof.

Page No 463:

Answer:

Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We know that the diagonals of a rhombus bisect each other at right angles.
$∴$ If AC = 24 cm and BD =10 cm, AO = 12 cm and BO = 5 cm
Applying Pythagoras theorem in right-angled triangle AOB, we get:
$A B^{2} = A O^{2} + B O^{2} = 12^{2} + 5^{2} = 144 + 25 = 169 A B = 13 cm$

Hence, the length of each side of the given rhombus is 13 cm.

Page No 463:

Answer:

Let the two triangles be ABC and PQR.
We have:
$△ A B C ~ △ P Q R$ ,
Here,
BC = a, AC = b and AB = c
PQ = r, PR = q and QR = p

We have to prove:
$\frac{a}{p} = \frac{b}{q} = \frac{c}{r} = \frac{a + b + c}{p + q + r}$

$△ A B C ~ △ P Q R$ ; therefore, their corresponding sides will be proportional.
$\Rightarrow \frac{a}{p} = \frac{b}{q} = \frac{c}{r} = k (say) . . . (i) \Rightarrow a = k p, b = k q and c = k r ∴ \frac{Perimeter of △ A B C}{Perimeter of △ P Q R} = \frac{a + b + c}{p + q + r} = \frac{k p + k q + k r}{p + q + r} = k . . . (ii) From (i) and (ii), we get : \frac{a}{p} = \frac{b}{q} = \frac{c}{r} = \frac{a + b + c}{p + q + r} = \frac{Perimeter of △ A B C}{Perimeter of △ P Q R}$

This completes the proof.

Page No 464:

Answer:

$Construction : Draw AX ⊥ CO and DY ⊥ BO . As, \frac{ar (△ ABC)}{ar (△ DBC)} = \frac{\frac{1}{2} \times AX \times BC}{\frac{1}{2} \times DY \times BC} \Rightarrow \frac{ar (△ ABC)}{ar (△ DBC)} = \frac{AX}{DY} . . . (i) In △ ABC and △ DBC, ∠ AXY = ∠ DYO = 90 ° (By construction) ∠ AOX = ∠ DOY (Vertically opposite angles) ∴ △ AXO \sim △ DYO (By AA criterion) ∴ \frac{AX}{DY} = \frac{AO}{DO} (Thales' s theorem) . . . (ii) From (i) and (ii), we have : \frac{ar (△ ABC)}{ar (△ DBC)} = \frac{AX}{DY} = \frac{AO}{DO} or, \frac{ar (△ ABC)}{ar (△ DBC)} = \frac{AO}{DO}$
This completes the proof.

Page No 464:

Answer:

$In △ A B C and △ B X Y, we have : ∠ B = ∠ B ∠ B X Y = ∠ B A C (Corresponding angles) Thus, △ A B C ~ △ B X Y (A A c r i t e r i o n) ∴ \frac{a r (△ A B C)}{a r (△ B X Y)} = \frac{A B^{2}}{B X^{2}} = \frac{A B^{2}}{{(A B - A X)}^{2}} . . . (i) Also, \frac{a r (△ A B C)}{a r (△ B X Y)} = \frac{2}{1} {∵ a r (△ B X Y) = a r (t r a p e z i u m A X Y C)} . . . (ii) From (i) and (ii), we have : \frac{A B^{2}}{{(A B - A X)}^{2}} = \frac{2}{1} \Rightarrow \frac{A B}{(A B - A X)} = \sqrt{2} \Rightarrow \frac{(A B - A X)}{A B} = \frac{1}{\sqrt{2}} \Rightarrow 1 - \frac{A X}{A B} = \frac{1}{\sqrt{2}} \Rightarrow \frac{A X}{A B} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} = \frac{(2 - \sqrt{2})}{2}$

Page No 464:

Answer:

Applying Pythagoras theorem in right-angled triangle ADC, we get:

$A C^{2} = A D^{2} + D C^{2} \Rightarrow A C^{2} - D C^{2} = A D^{2} \Rightarrow A D^{2} = A C^{2} - D C^{2} . . . (1)$

Applying Pythagoras theorem in right-angled triangle ADB, we get:

$A B^{2} = A D^{2} + D B^{2} \Rightarrow A B^{2} - D B^{2} = A D^{2} \Rightarrow A D^{2} = A B^{2} - D B^{2} . . . (2)$

$From equation (1) and (2), we have : A C^{2} - D C^{2} = A B^{2} - D B^{2} \Rightarrow A C^{2} = A B^{2} + D C^{2} - D B^{2} \Rightarrow A C^{2} = A B^{2} + (D B + B C)^{2} - D B^{2} (∵ D B + B C = D C) \Rightarrow A C^{2} = A B^{2} + D B^{2} + B C^{2} + 2 D B . B C - D B^{2} \Rightarrow A C^{2} = A B^{2} + B C^{2} + 2 B C . B D$

This completes the proof.

Page No 464:

Answer:

$In △ P A C and △ Q B C, we have : ∠ A = ∠ B (Both angles are 90 °) ∠ P = ∠ Q (Correspond ing angles) and ∠ C = ∠ C (Common angles) Therefore, △ P A C ~ △ Q B C \frac{A P}{B Q} = \frac{A C}{B C}$

$\Rightarrow \frac{x}{z} = \frac{a + b}{b} \Rightarrow a + b = \frac{b x}{z} . . . (1)$

$In △ R C A and △ Q B A, we have : ∠ C = ∠ B (Both angles are 90 °) ∠ R = ∠ Q (Correspond ing angles) and ∠ A = ∠ A (Common angles) Therefore, △ R C A ~ △ Q B A \frac{R C}{B Q} = \frac{A C}{A B} \Rightarrow \frac{y}{z} = \frac{a + b}{a} \Rightarrow a + b = \frac{a y}{z} . . . (2)$
From equation (1) and (2), we have:

$\frac{b x}{z} = \frac{a y}{z} \Rightarrow b x = a y \Rightarrow \frac{a}{b} = \frac{x}{y} . . . (3) Also, \frac{x}{z} = \frac{a + b}{b} \Rightarrow \frac{x}{z} = \frac{a}{b} + 1 Using the value of \frac{a}{b} from equation (3), we have : \Rightarrow \frac{x}{z} = \frac{x}{y} + 1 Dividing both sides by x, we get : \frac{1}{z} = \frac{1}{y} + \frac{1}{x}$

$∴ \frac{1}{x} + \frac{1}{y} = \frac{1}{z} This completes the proof .$

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