Rd Sharma 2022 for Class 10 Maths Chapter 7 - Triangles

Answer:

We have an equilateral triangle in which AD is altitude. An equilateral triangle is drawn using AD as base. We have to prove that,

Since the two triangles are equilateral, the two triangles will be similar also.

We know that according to the theorem, the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

…… (1) 

Nowis an equilateral triangle. So,

.

Therefore,

So,

We will now use this in equation (1). So,

Hence, proved.

Answer:

(i) Since all circles have centre and circumference, therefore all circles are similar.

Hence

(ii) Since all squares have each angle and sides are proportional, therefore all squares are similar.

Hence

(iii) In equilateral triangle each angle is therefore all equilateral triangles are similar.

Hence

(iv) Two triangles are similar, if their corresponding angles are _.

(v) Two triangles are similar, if their corresponding sides are.

(vi) Two polygons of same number of sides are similar, if (a) their corresponding angles are and (b) their corresponding sides are .

Answer:

(i)

(ii)

(iii)

(iv)

(v)

(vi)
 

Answer:

For two figures to be similar, the ratio of their corresponding sides must be equal.

Let the other rectangle be ABCD with AB = 6 cm and AD = 3.9 cm.

Here,
$\frac{\mathrm{PQ}}{\mathrm{AB}}=\frac{4}{6}=\frac{2}{3}$

And
$\frac{\mathrm{PS}}{\mathrm{AD}}=\frac{2.6}{3.9}=\frac{2}{3}$

Since $\frac{\mathrm{PQ}}{\mathrm{AB}}=\frac{\mathrm{PS}}{\mathrm{AD}}=\frac{2}{3}$, the two rectangles are similar.

Answer:

(i) It is given that  and  DE || BC

We have to find the

Since 

So  (by Thales theorem)

Then 

Hence

(ii) It is given that  and 

We have to find 

Let 

So  (by Thales theorem)

Then 

Hence

(iii)  It is given that , and .

So       (by Thales theorem)

Then 

Now

(iv)It is given that ,, and .

We have to find 

So  (by Thales theorem)

Then 

Hence

(v) It is given that , and .

We have to find CE.

So      (by Thales theorem)

Then 

Hence

(vi)

It is given that ,, and .

We have to find BD and CE.

Since DE$\parallel$BC, AB is transversal, then

∠ADE = ∠ABC   (corresponding angles)

Since DE$\parallel$BC, AC is a transversal, then

∠AED = ∠ACB   (corresponding angles)

In ∆ADE and ∆ABC,

∠ADE = ∠ABC  (proved above)

∠AED = ∠ACB  (proved above)

so, ∆ADE ∼ ∆ABC (Angle Angle similarity)

Since, the corresponding sides of similar triangles are proportional, then

$$\begin{gathered} \frac{\mathrm{AD}}{\mathrm{AB}} = \frac{\mathrm{AE}}{\mathrm{AC}} = \frac{\mathrm{DE}}{\mathrm{BC}} \\ \Rightarrow \frac{\mathrm{AD}}{\mathrm{AB}} = \frac{\mathrm{DE}}{\mathrm{BC}} \\ \Rightarrow \frac{2.4}{2.4 + \mathrm{DB}} = \frac{2}{5} \\ \Rightarrow 2.4 + \mathrm{DB} = 6 \\ \Rightarrow \mathrm{DB} = 6 - 2.4 \\ \Rightarrow \mathrm{DB} = 3.6 \mathrm{cm} \\ \mathrm{Similarly}, \\ \frac{\mathrm{AE}}{\mathrm{AC}} = \frac{\mathrm{DE}}{\mathrm{BC}} \\ \Rightarrow \frac{3.2}{3.2 + \mathrm{EC}} = \frac{2}{5} \\ \Rightarrow 3.2 + \mathrm{EC} = 8 \\ \Rightarrow \mathrm{EC} = 8 - 3.2 \\ \Rightarrow \mathrm{EC} = 4.8 \mathrm{cm} \end{gathered}$$

Hence, BD = 3.6 cm  and  CE = 4.8 cm.

Answer:

(i) It is given that  and  are point on sides AB and AC.

We have to prove that DE || BC.

According to Thales theorem we have

$\Rightarrow \frac{AD}{AB-AD}=\frac{AE}{AC-AE}$

        (Proportional)

Hence, DE || BC.

(ii) It is given that  and  are point on sides AB and AC.

We have to prove that DE || BC.

According to Thales theorem we have

    (Proportional)

Hence, DE || BC.

Answer:

It is given that EP = 3 cm, PG = 3.9 cm, FQ = 3.6 cm and FQ = 2.4 cm.

We have to check that or not.

According to Thales theorem we have

$\frac{PG}{GE}=\frac{GQ}{FQ}$

Now,

$\frac{3.9}{3}\ne \frac{3.6}{2.4}$

Hence, it is not proportional.

So, PQ ∦ EF.

Answer:

By the Basic Proportionality Theorem, "If a line is drawn parallel to one side of a triangle intersects the other two sides in distinct points, then the other two sides are divided in the same ratio."

For DE $\parallel$ AB, the ratio in which they divide the sides should be equal.

Thus,
$$\begin{gathered} \frac{\mathrm{CD}}{\mathrm{AD}}=\frac{\mathrm{CE}}{\mathrm{BE}} \\ \Rightarrow \frac{x+3}{3x+19}=\frac{x}{3x+4} \\ \Rightarrow \left(x+3\right)\left(3x+4\right)=x\left(3x+19\right) \\ \Rightarrow 3x^2+9x+4x+12=3x^2+19x \\ \Rightarrow 13x+12=19x \\ \Rightarrow 6x=12 \\ \Rightarrow x=2 \end{gathered}$$

Hence, $x=2$.

Answer:

Given that, AB || CD.

Consider $△$AOB and $△$COD.
$$\begin{gathered} \angle \mathrm{OBA}=\angle \mathrm{ODC} \left(\mathrm{vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle \mathrm{OAB}=\angle \mathrm{OCD} \left(\mathrm{vertically} \mathrm{opposite} \mathrm{angles}\right) \end{gathered}$$

Therefore, by AA similarity criterion, $△$AOB ∼ $△$COD.

Since the two triangles are similar, the ratio of their corresponding sides is equal.
Thus,
$$\begin{gathered} \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}} \\ \Rightarrow \frac{3x-1}{5x-3}=\frac{2x+1}{6x-5} \\ \Rightarrow \left(3x-1\right)\left(6x-5\right)=\left(5x-3\right)\left(2x+1\right) \\ \Rightarrow 18x^2-21x+5=10x^2-x-3 \\ \Rightarrow 8x^2-20x+8=0 \\ \Rightarrow 8x^2-16x-4x+8=0 \\ \Rightarrow 8x\left(x-2\right)-4\left(x-2\right)=0 \\ \Rightarrow \left(x-2\right)\left(8x-4\right)=0 \\ \Rightarrow x=2 \mathrm{or} x=\frac{1}{2} \end{gathered}$$

For $x=\frac{1}{2}$, OD = $6x-5=6\times \frac{1}{2}-5=-2$ which is not possible

Hence, x = 2.

Answer:

Given that, AB || CD.

Consider $△$AOB and $△$COD.
$$\begin{gathered} \angle \mathrm{OBA}=\angle \mathrm{ODC} \left(\mathrm{vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle \mathrm{OAB}=\angle \mathrm{OCD} \left(\mathrm{vertically} \mathrm{opposite} \mathrm{angles}\right) \end{gathered}$$

Therefore, by AA similarity criterion, $△$AOB ∼ $△$COD.

Since the two triangles are similar, the ratio of their corresponding sides is equal.
Thus,
$$\begin{gathered} \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}} \\ \Rightarrow \frac{3x-19}{x-3}=\frac{x-4}{4} \\ \Rightarrow 4\left(3x-19\right)=\left(x-3\right)\left(x-4\right) \\ \Rightarrow 12x-76=x^2-7x+12 \\ \Rightarrow x^2-19x+88=0 \\ \Rightarrow x^2-8x-11x+88=0 \\ \Rightarrow x\left(x-8\right)-11\left(x-8\right)=0 \\ \Rightarrow \left(x-8\right)\left(x-11\right)=0 \\ \Rightarrow x=8 \mathrm{or} x=11 \end{gathered}$$

Hence, $x=8 \mathrm{or} x=11$.

Answer:

It is given that in , DE || BC and BD = CE.

We have to prove that ∆ABC is isosceles.

By Thales theorem we have





Now and

So

Hence

So, ∆ABC is isosceles

Answer:

(i) It is given that , and .

In , is the bisector of, meeting side  at .

We have to find .

Since is  bisector

Then 

Hence 

(ii) It is given that , and.

In , is the bisector of, meeting side  at .

We have to find .

Since is  bisector

So  (is bisector of  and side)

Then

Hence 

(iii) It is given that , and .

In , is the bisector of, meeting side  at .

We have to find BD and .

Since is  bisector

So 

Let BD = x cm

Then

⇒

Now 

Hence  and 

Answer:

It is given that is the bisector of the exterior

Meeting produced and, and  

Since is the bisector of the exterior

So

Hence

Answer:

It is given that in , , and .

We have to find .

In ,

Since , therefore, AD is the bisector of $\angle A$.

Hence,

Answer:

(i) It is given that,,, and .

We have to check whether is bisector of .

First we will check proportional ratio between sides.

So 
$$\begin{gathered} \Rightarrow \frac{4}{6}=\frac{1.6}{2.4} \\ \Rightarrow \frac{2}{3}=\frac{2}{3} \end{gathered}$$

(It is proportional)

Hence,  is bisector of .

(ii) It is given that,,, and .

We have to check whether is bisector of .

First we will check proportional ratio between sides.

Now

So 

(It is proportional)

Hence,  is bisector of .

Answer:

It is given that , and .

We have to find _, and BD.

Sinceis bisector of

So

Then,

$$\begin{gathered} \frac{5}{4}=\frac{BD}{BC-BD} \\ \Rightarrow \frac{5}{4}=\frac{BD}{8-BD} \\ \Rightarrow 40-5BD=4BD \\ \Rightarrow 9BD=40 \end{gathered}$$

So,

Since is the bisector of .

So,

$$\begin{gathered} \frac{AB}{BC}=\frac{AE}{EC} \\ \Rightarrow \frac{AB}{BC}=\frac{AC-EC}{EC} \end{gathered}$$

So

Now since is the bisector of

So

So 

Hence

And

Answer:

(i) It is given that  and _. 

When ,we have to find the.

Since is right angle triangle and  is perpendicular on , so

   (AA similarity)

⇒ DC = 16 cm

Hence, .

(ii) Given that, ∠ABC = 90°, BD ⊥ AC, AD = 4 cm and CD = 5 cm.
Let $\angle \mathrm{DCB}=x$.
$$\begin{gathered} \Rightarrow \angle \mathrm{DBC}=90°-x \\ \Rightarrow \angle \mathrm{DBA}=x \end{gathered}$$

Consider $△$ADB and $△$BDC. 
$$\begin{gathered} \angle \mathrm{B}=\angle \mathrm{D}=90° \\ \angle \mathrm{DBC}=\angle \mathrm{DBA}=x \end{gathered}$$

Therefore, by AA similarity criterion, $△$ABD $\approx$ $△$BDC.

Thus,
$$\begin{gathered} \frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{BD}}{\mathrm{DC}} \\ \Rightarrow {\mathrm{BD}}^2=\mathrm{AD}·\mathrm{CD} \\ \Rightarrow {\mathrm{BD}}^2=4·5 \\ \Rightarrow {\mathrm{BD}}^2=20 \\ \Rightarrow \mathrm{BD}=2\sqrt{5} \mathrm{cm} \end{gathered}$$

Now, in $△$ABC,
$$\begin{gathered} {\mathrm{AB}}^2={\mathrm{BD}}^2+{\mathrm{AD}}^2 \\ \Rightarrow {\mathrm{AB}}^2={\left(2\sqrt{5}\right)}^2+4^2 \\ \Rightarrow {\mathrm{AB}}^2=20+16 \\ \Rightarrow {\mathrm{AB}}^2=36 \\ \Rightarrow \mathrm{AB}=6 \mathrm{cm} \left(\because \mathrm{side} \mathrm{cannot} \mathrm{be} \mathrm{negative}\right) \end{gathered}$$

(iii) It is given that,,,  and 

We have to find .

Since       

So 

Hence, 

Answer:

It is given that .

,, and.

We have to find and .

Since

So 

Similarly

Hence, and

Answer:

It is given that

,and PR = 6 cm

We have to find .

Since

So 

Hence,

Answer:

It is given that .

,and

We have to find .

Since

()

So 

Hence,

Answer:

It is given that, and .

We have to find AD.

Since

So 

Hence,

Answer:

Given: $\mathrm{AB}\perp \mathrm{BC}, \mathrm{DC}\perp \mathrm{BC} \mathrm{and} \mathrm{DE}\perp \mathrm{AC}$
To prove: $∆\mathrm{CED}~∆\mathrm{ABC}$
Proof:
Consider the figure.
In $∆\mathrm{DEC},$
$$\begin{gathered} \angle \mathrm{DEC}={90}^{\circ } \\ \mathrm{Also}, \angle \mathrm{DEC}+\angle \mathrm{ECD}+\angle \mathrm{CDE}={180}^{\circ } (\mathrm{Angle} \mathrm{sum} \mathrm{property} \mathrm{of} \mathrm{triangle}) \\ \Rightarrow {90}^{\circ }+\angle \mathrm{ECD}+\angle \mathrm{CDE}={180}^{\circ } \\ \Rightarrow \angle \mathrm{ECD}+\angle \mathrm{CDE}={90}^{\circ } ...\left(\mathrm{i}\right) \end{gathered}$$
Also given that $\mathrm{DC}\perp \mathrm{BC}$
$$\begin{gathered} \Rightarrow \angle \mathrm{DCB}={90}^{\circ } \\ \Rightarrow \angle \mathrm{ECD}+\angle \mathrm{ECB}={90}^{\circ } ...\left(\mathrm{ii}\right) \end{gathered}$$
Comparing (i) and (ii), we get
$$\begin{gathered} \angle \mathrm{ECD}+\angle \mathrm{ECB}=\angle \mathrm{ECD}+\angle \mathrm{CDE} \\ \Rightarrow \angle \mathrm{ECB}=\angle \mathrm{CDE} ...\left(\mathrm{iii}\right) \end{gathered}$$ 
Also because $\mathrm{AB}\perp \mathrm{BC} \mathrm{and} \mathrm{DC}\perp \mathrm{BC}$,
$\Rightarrow \mathrm{AB}\parallel \mathrm{DC}$ and AC acts as a transversal.
$\Rightarrow \angle \mathrm{BAC}=\angle \mathrm{ECD}$ (Alternate angles)                    ...(iv)
Again, $\angle \mathrm{ABC}=\angle \mathrm{CED} (\mathrm{Both} {90}^{\circ }) ...\left(\mathrm{v}\right)$
From (iii), (iv) and (v) we have 
$△\mathrm{ABC}~△\mathrm{CED}$ (AAA similarity criteria)

Answer:

It is given that trapezium with . O is the point of intersection of AC  and BD.

We have to prove that

Now, in and

$\angle AOB=\angle COD$    (Vertically opposite angles)

$\angle OAB=\angle OCD$    (Alternate angles)

∴  $~$        (AA Similarity)

Hence,                 (Corresponding sides are proportional)

Answer:

Let ∆ABC be a right angle triangle having sides and ; and hypotenuse . BD is the altitude drawn on the hypotenuse AC.

We have to find to prove .

Since the altitude is perpendicular on the hypotenuse, both the triangles are similar

Hence, .

Answer:

Comparing ΔCAB and ΔCED,

∠CAB = ∠CED                [Given]

∠ACB = ∠ECD                [Common]

∴ ΔCAB ∼ ΔCED

Answer:

It is given that is right angle triangle and

We have to prove that and find the lengths of and .

In ∆ABC ∼ ∆ADE,

$$\begin{gathered} \angle A=\angle A \left(\mathrm{Common}\right) \\ \angle C=\angle E \left(90°\right) \end{gathered}$$

So by similarly criterion, we have

Since

So

And

Hence, and

Answer:

It is given that perimeter of two similar triangle are and and one side.

We have to find the other side.

Let the corresponding side of the other triangle be x cm.

$\frac{25 \mathrm{cm}}{15 \mathrm{cm}}=\frac{9 \mathrm{cm}}{x}$

Hence

Answer:

It is given that, girl height, speed and height of lamp.

We have to find the length of her shadow after

Let be the lamp post and be the girl.

Suppose is the length of her shadow.
Let DE = x

And 

Now in and we have

and

So by similarly criterion

$$\begin{gathered} \frac{BE}{DE}=\frac{AB}{CD} \\ \frac{4.8+x}{x}=\frac{3.6}{0.9}=4 \\ \Rightarrow 3x=4.8 \\ \Rightarrow x=1.6 \end{gathered}$$

Hence the length of her shadow after is 1.6 m.

Answer:

It is given that length of vertical stick

We have to find the height of the tower.

Suppose is the height of the tower and is its shadow.

Now,      $\left(\angle B=\angle Q \mathrm{and} \angle A=\angle P\right)$

Hence the height of the tower is _.

Answer:

It is given that _, and are each perpendicular to .

We have to prove that

In we have

$\Rightarrow \frac{y}{x}=\frac{CB}{CA}$  ......(1)

Now in _, we have

…… (2)

Adding (1) and (2) we have

Hence, .

Answer:

It is given that .

,and

We have to calculate the values of and .

In and , we have

         (Vertically opposite angles)

     (Alternate interior angles)

So
$\frac{EF}{AB}=\frac{DE}{DB}$

Similarly in we have
$$\begin{gathered} \frac{DC}{AB}=\frac{DE}{BE} \\ \Rightarrow \frac{x}{6}=\frac{y}{4+y} \\ \Rightarrow x=\frac{6y}{4+y} \\ \Rightarrow x=\frac{6\times 6.67}{4+6.67} \\ \Rightarrow x=3.75 \end{gathered}$$

Hence, $x=3.75\mathrm{cm}$ and

Answer:

(i) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

(ii) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Answer:

Given: ΔACB is similar to ΔAPQ.

BC = 10 cm, PQ = 5cm, BA = 6.5cm and AP = 2.8 cm 

TO FIND:

(1) CA and AQ

(2) Area of ΔACB : Area of ΔAPQ

(1) It is given that ΔACB $~$ ΔAPQ.

We know that for any two similar triangles the sides are proportional. Hence 

$\frac{\mathrm{AB}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{AP}}$

Similarly,

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

$\frac{\mathrm{ar}\left(∆\mathrm{ACQ}\right)}{\mathrm{ar}\left(∆\mathrm{APQ}\right)}={\left(\frac{\mathrm{BC}}{\mathrm{PQ}}\right)}^2={\left(\frac{10}{5}\right)}^2={\left(\frac{2}{1}\right)}^2=\frac{4}{1}$

Answer:

In the given figure, we have DE || BC.

$\angle ADE=\angle B \left(\mathrm{Corresponding} \mathrm{angles}\right)$

$\angle DAE=\angle BAC \left(\mathrm{Common}\right)$

So, $∆ADE~∆ABC$      (AA Similarity)

(i) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Hence 

(ii) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Hence,

(iii) We know that 

Let Area of ΔADE = 9x sq. units and Area of ΔABC = 25x sq. units

Now,

(iv) Given that, DE || BC, DE = 6 cm and BC = 12 cm.
We already have that $△\mathrm{ADE}~△\mathrm{ABC}$. 

Thus,
$\begin{array}{rcl}\frac{\mathrm{ar}\left(△\mathrm{ADE}\right)}{\mathrm{ar}\left(△\mathrm{ABC}\right)}& =& \frac{6^2}{{12}^2}\\ & =& \frac{1}{4}\end{array}$

So, if $\begin{array}{rcl}\mathrm{ar}\left(△\mathrm{ABC}\right)& =& 4k\end{array}$, then $\begin{array}{rcl}\mathrm{ar}\left(△\mathrm{ADE}\right)& =& k\end{array}$.
$\begin{array}{rcl}& \Rightarrow & \mathrm{ar}\left(\mathrm{quad}. \mathrm{DECB}\right)=3k\end{array}$

Thus,
$\begin{array}{rcl}\frac{\mathrm{ar}\left(△\mathrm{ADE}\right)}{\mathrm{ar}\left(\mathrm{quad}. \mathrm{DECB}\right)}& =& \frac{k}{3k}\\ & =& \frac{1}{3}\end{array}$

Answer:

Given: The area of two similar triangles is 81cm² and 49cm² respectively.

To find:

(1) Ratio of their corresponding heights.

(2) Ratio of their corresponding medians.

(1) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Taking square root on both sides, we get

(2) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their medians.

Taking square root on both sides, we get

Answer:

Given: The area of two similar triangles is 169cm² and 121cm² respectively.The longest side of the larger triangle is 26cm.

To find: Longest side of the smaller triangle

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

Taking square root on both sides, we get

= 22 cm

Hence, the longest side of the smaller triangle is .

Answer:

Given: The area of two similar triangles is 25cm² and 36cm² respectively. If the altitude of first triangle is 2.4cm

To find: The altitude of the other triangle 

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Taking square root on both sides, we get

$$\begin{gathered} \frac{5}{6}=\frac{2.4}{\mathrm{altitude}2} \\ \Rightarrow \mathrm{altitude}2=2.88 \mathrm{cm} \end{gathered}$$

Hence, the corresponding altitude of the other is 2.88 cm.

Answer:

Given: The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively.

To find: Ratio of areas of triangle.

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Hence, the ratio of areas of two triangles is 4 : 9.

Answer:

Given: The area of two similar triangles is 100cm² and 49cm² respectively. If the altitude of bigger triangle is 5 cm

To find: their corresponding altitude of other triangle

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Taking square root on both side

Answer:

Given: The area of two similar triangles is 121cm² and 64cm² respectively. IF the median of the first triangle is 12.1cm

To find: corresponding medians of the other triangle

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their medians.

Answer:

Given: The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. BC = 4.5cm.

To find: length of QR 

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

$\frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{PQR}\right)}={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^2$
           

Answer:

Given: In ΔABC, PQ is a line segment intersecting AB at P, and AC at Q. AP = 1cm, PB = 3cm, AQ = 1.5cm and QC = 4.5cm.

To find:

In ΔABC,

According to converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Hence, PQ || BC

In ΔAPQ and ΔABC,

$\angle \mathrm{APQ}=\angle \mathrm{B} \left(\mathrm{Corresponding} \mathrm{angles}\right)$

$\angle \mathrm{PAQ}=\angle \mathrm{BAC} \left(\mathrm{Common}\right)$

So, $∆\mathrm{APQ}~∆\mathrm{ABC}$      (AA Similarity)

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

Hence 

Answer:

Given: In ΔABC, D is a point on side AB such that AD : DB= 3 : 2. E is a point on side BC such that DE || AC.

To find:

In ΔABC,

According to converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Hence DE || AC

In ΔBDE and ΔABC,

$\angle \mathrm{BDE}=\angle \mathrm{A} \left(\mathrm{Corresponding} \mathrm{angles}\right)$

$\angle \mathrm{DBE}=\angle \mathrm{ABC} \left(\mathrm{Common}\right)$

So, $∆\mathrm{BDE}~∆\mathrm{ABC}$      (AA Similarity)

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

Let AD = 2x and BD = 3x.

Hence

Answer:

Given: In ΔABC and ΔBDE are equilateral triangles. D is the midpoint of BC.

To find: $\frac{\mathrm{Ar}\left(∆\mathrm{ABC}\right)}{\mathrm{Ar}\left(∆\mathrm{BDE}\right)}$

In ΔABC and ΔBDE

Since D is the midpoint of BC, BD : DC = 1.

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

Let DC = x, and BD = x

Therefore BC = BD + DC = 2x

Hence 

Answer:

It is given that the diagonals of the trapezium PQRS intersect each other at the point O. Also, PQ || RS and PQ = 3RS.

In ∆SOR and ∆QOP,

∠SOR = ∠QOP      (Vertically opposite angles)

∠OSR = ∠OQP       (Alternate angles)

∴ ∆SOR $~$ ∆QOP    (AA Similarity)

$\Rightarrow \frac{\mathrm{ar}\left(∆\mathrm{POQ}\right)}{\mathrm{ar}\left(∆\mathrm{ROS}\right)}={\left(\frac{\mathrm{PQ}}{\mathrm{RS}}\right)}^2$       (The ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides)

$\Rightarrow \frac{\mathrm{ar}\left(∆\mathrm{POQ}\right)}{\mathrm{ar}\left(∆\mathrm{ROS}\right)}={\left(\frac{3\mathrm{RS}}{\mathrm{RS}}\right)}^2=\frac{9}{1}$

Hence, the ratio of areas of triangles POQ and ROS is 9 : 1.

Answer:

Given: ΔABC and ΔDBC are on the same base BC. AD and BC intersect at O.

Prove that:

Construction: Draw and .

Now, in ΔALO and ΔDMO, we have 

$\therefore \frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}$                (Corresponding sides are proportional)

$$\begin{gathered} \frac{\mathrm{Ar}\left(∆\mathrm{ABC}\right)}{\mathrm{Ar}\left(∆\mathrm{BCD}\right)}=\frac{{\displaystyle \frac{1}{2}}\times \mathrm{BC}\times \mathrm{AL}}{{\displaystyle \frac{1}{2}}\times \mathrm{BC}\times \mathrm{DM}} \\ =\frac{{\displaystyle \mathrm{AL}}}{\mathrm{DM}} \\ =\frac{\mathrm{AO}}{\mathrm{DO}} \end{gathered}$$

Answer:

(i) Let

In order to prove that the given sides of a certain triangle forms a right angled triangle we have to prove that square of the larger side is equal to the sum of the squares of the other two sides.

Here, the larger side is.

Hence, we have to prove that.

Let solve the left hand side of the above equation.

Now we will solve the right hand side of the equation,

Here we can observe that left hand side is equal to the right hand side that is .

Therefore, the given sides of a certain triangle form a right angled triangle.

(ii) Let

In order to prove that the given sides of a certain triangle forms a right angled triangle we have to prove that square of the larger side is equal to the sum of the squares of the other two sides.

Here, the larger side is .

Hence, we have to prove that.

Let solve the left hand side of the above equation.

Now we will solve the right hand side of the equation,

Here we can observe that left hand side is not equal to the right hand side.

Therefore, the given sides of a certain triangle do not form a right angled triangle.

(iii) Let

In order to prove that the given sides of a certain triangle forms a right angled triangle we have to prove that square of the larger side is equal to the sum of the squares of the other two sides.

Here, the larger side is .

Hence, we have to prove that.

Let solve the left hand side of the above equation.

Now we will solve the right hand side of the equation,

Here we can observe that left hand side is not equal to the right hand side.

Therefore, the given sides of a certain triangle do not form a right angled triangle.

(iv) Let

In order to prove that the given sides of a certain triangle forms a right angled triangle we have to prove that square of the larger side is equal to the sum of the squares of the other two sides.

Here, the larger side is .

Hence, we have to prove that .

Let solve the left hand side of the above equation.

Now we will solve the right hand side of the equation,

Here we can observe that left hand side is equal to the right hand side that is .

Therefore, the given sides of a certain triangle form a right angled triangle.

Answer:

Let us draw the diagram from the given information we get a right angled triangle ABC as shown below,

Let the window be at the point A. We know that angle formed between the building and ground is always 90°.

Given: AB = 15 m and CA = 17 m

Now we will use Pythagoras theorem to find .

Let us substitute the values we get,

Subtracting 225 from both the sides of the equation we get,

Let us take the square root we get,

Therefore, the distance of the foot of the ladder from the building is.

Answer:

Let us draw the diagram from the given information.

Let us draw a perpendicular from B on CD which meets CD at P.

It is clear that BP = 12 m because it is given that distance between feet of the two poles is 12 m.

After drawing the perpendicular we get a rectangle BACP such that AB = PC and BP = AC.

Because of this construction we also obtained a right angled triangle BPD.

Now we will use Pythagoras theorem,

Let us substitute the values of BP and PD we get,

Taking the square root we get,

Therefore, distance between the top of the two poles is .

Answer:

We know that altitude that is a perpendicular drawn on the unequal side of the isosceles triangle bisects that side.

Therefore, BD = DC = 7 cm.

Let us use the Pythagoras theorem in right angled triangle ADB we get,

Substituting the values we get,

Subtracting 49 from both the sides we get,

Let us take the square root we get,

AD = 24 cm

Therefore, the altitude of the isosceles triangle is .

Answer:

The given information can be represented as follows.

Here, A is the position of the window and AC is the ladder.

Also, DE is the same ladder when it is shifted.

C and E are the original and final position of the foot of the ladder.

Now, applying Pythagoras theorem in ΔABC,

AC² = AB² + BC²

⇒ AC² = (8 m)² + (6 m)² = (10 m)²

⇒ AC = 10 m

Now, again applying Pythagoras theorem in ΔEBD

DE² = EB² + BD²

⇒ (10 m)² = (8 m)² + BD²

⇒ BD² = 100 m² − 64 m² = 36 m²

⇒ BD = 6 m

Thus, the tip of the ladder is now at the height of 6 m above the ground.

Answer:

Let us draw the figure first.

An aeroplane which flies due north at a speed of 300 km/hr covers the distance AB after $1\frac{1}{2} \mathrm{hours}$ and another aeroplane that flies due west at the speed of 400 km/hr covers the distance BC after $1\frac{1}{2} \mathrm{hours}$.

We know that 

Let us calculate AB first as shown below,

AB = 300 × 1.5 km

⇒ AB = 450 km

Similarly, we can calculate BC.

BC = 400 × 1.5 km

⇒ BC = 600 km

Now we have to find AC. To find AC we will use Pythagoras theorem,

$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2 \\ \Rightarrow {\mathrm{AC}}^2={\left(300\right)}^2+{\left(400\right)}^2 \\ \Rightarrow {\mathrm{AC}}^2=90000+160000 \\ \Rightarrow {\mathrm{AC}}^2=250000 \end{gathered}$$

Taking square root we get,

AC = 500

Therefore, after $1\frac{1}{2} \mathrm{hours}$ the aeroplanes will be approximately 500 km far apart.

Answer:

In and

$\angle B=\angle C$        (60º each)

$\angle ADB=\angle ADC$           (90º each)

    (AAS congruence theorem)

But therefore, we get,

………(1)

Now we will divide both sides of the equation (1) by 2, we get,

Now we will use Pythagoras theorem in right angled triangle ADB.

Now we will substitute the values of AB and BD we get,

Therefore, .

We have given an equilateral triangle and we know that the area of the equilateral triangle is.

Here, side is 2a

Therefore, .

Answer:

It is given that F is the midpoint of AB. Therefore, we have AF = FB.

It is also given that               .....(1)

Now look at the figure. Quadrilateral ABCD is a square and hence all angles are of 90º.

In , and hence it is a right angle triangle.

We know that the area of the right angle triangle is $\frac{1}{2}$ × base  × height

Therefore, $Ar\left(∆FBE\right)=\frac{1}{2}\times BF\times BE$

        cm²

Now we will multiple both sides of the equation by 2 we get,                   .....(2)

But we know that and.

Let us substitute these values in equation (2) we get,

Let us simplify the above equation as below,

But we know that ABCD is a square and hence AB = BC = CD = AD.

                .....(3)

We know that 216 is the cube of 6 therefore we can write the equation (3) as below,

AB² = 6³ × 6

AB² = 6⁴

Therefore, side of the square ABCD is 36 cm.

Now we are going to find the diagonal AC.

Diagonal of the square can be calculate by using the formula given below,

Diagonal = $\sqrt{2}$ Side

                         .....(4)

We know that

Let us substitute the value of in equation (3).

Therefore, the length of AC is .

Answer:

We know that altitude that is a perpendicular drawn on the unequal side of the isosceles triangle bisects that side.

Therefore, BD = DC = 7 cm.

Let us use the Pythagoras theorem in right angled triangle ADB we get,

Substituting the values we get,

Subtracting 49 from both the sides we get,

Let us take the square root we get,

AD = 24 cm

Therefore, the altitude of the isosceles triangle is .

Answer:

We will draw the figure from the given information as below,

Let AB be the vertical pole of length 18 m and let the stake be at the point C so the wire will be taut.

Therefore, we have , and we have to find BC.

Now we will use Pythagoras theorem,

Let us substitute the values we get,

Subtracting 324 from both sides of the equation we get,

We can rewrite the 252 as , therefore, our equation becomes,

Now we will take the square root,

Therefore, the stake should be far from the base of the pole so that the wire will be taut.

Answer:

We know that a quadrilateral is said to a rhombus if all sides of the quadrilateral are equal. Diagonals of a rhombus bisect each other at right angles.

Quadrilateral ABCD is a rhombus and diagonals AC and BD intersect at point O.

As we defined above, we get , , and angle .

We are given that AC = 10 cm and BD = 24 cm.

Therefore, we get, AO = OC = 5 cm and BO = OD = 12 cm.

Now we will use Pythagoras theorem in the right angled triangle AOD as below,

…......(1)

Now we will substitute the values of AO and OD in equation (1) we get,

Let us take the square root

AD = 13

Therefore, length of the side of the rhombus is .

Answer:

We know that a quadrilateral is said to a rhombus if all sides of the quadrilateral are equal. Diagonals of a rhombus bisect each other at right angles.

Quadrilateral ABCD is a rhombus and diagonals AC and BD intersect at point O.

As we defined above, we get , , and angle .

We are given that AB = 10 cm and AC = 16 cm. Now we will find length of BD.

As we know the definition of rhombus we get AB = BC = CD = AD.

Therefore, we get, AB = BC = CD = AD = 10 cm

Also we know that diagonals of rhombus bisect each other at right angles therefore, we get,

, 

and

Here, we know the length of AC therefore, we get, .

Now we will use Pythagoras theorem in the right angled triangle AOD as below,

…......(1)

Now we will substitute the values of AD and AO in equation (1) we get,

…......(2)

Now we will subtract 64 from both sides of the equation (2)

Now we will take the square root.

OD = 6

We know that

Therefore, length of the other diagonal is .

Answer:

We are asked to find the height of the equilateral triangle.

Let us draw the figure. Let us draw the altitude AD. We know that altitude is also median of the equilateral triangle.

Therefore,.

In right angled triangle ADB, we will Pythagoras theorem, as shown below,

Now we will substitute the values.

144 = AD²  + 36

AD² = 144 − 36 = 108

AD = 10.39 cm

Therefore, the height of the equilateral triangle is.

Answer:

Let

Larger side is

We know that any number plus 1 is always greater than that number minus 1 and product of 2 and its square root.

For example : If a = 36

If a = 5

In order to prove that the given sides forms a right angled triangle we have to prove that .

Let us solve the left hand side first.

Now we will simplify the right hand side as shown below,

We can see that left hand side is equal to right hand side.

Therefore, the given sides determined the right angled triangle.

Answer:

We have to prove that .

In right angled, using Pythagoras theorem we get,

….....(1)

We know that in an equilateral triangle every altitude is also median.

Therefore, AD bisects BC.

Therefore, we have

Since is an equilateral triangle,

Therefore, we can write equation (1) as

…......(2)

But

Therefore, equation (2) becomes,

Simplifying the equation we get,

Therefore, .

Answer:

(i)

In and ,

$\angle ACB=\angle A=90°$

(Common angle)

So, by AA criterion

                 .....(1)

(ii) In and ,

    (Common angle)

So, by AA criterion

             .....(2)

(iii) We have shown that is similar to and is similar to therefore, by the property of transitivity is similar to .

      .....(3)

(iv) Now to obtained AB²/AC² = BD/DC, we will divide equation (1) by equation (2) as shown below,

Canceling BC we get,

Therefore,

Answer:

(i) Since AD perpendicular to BC we obtained two right angled triangles, triangle ADB and triangle ADC.

We will use Pythagoras theorem in the right angled triangle ADC

…......(1)

Let us substitute AD = h, AC = b and DC = (a − x) in equation (1) we get,

…......(2)

(ii) Let us use Pythagoras theorem in the right angled triangle ADB as shown below,

…......(3)

Let us substitute AB = c, AD = h and BD = x in equation (3) we get,

Let us rewrite the equation (2) as below,

…......(4)

Now we will substitute in equation (4) we get,

Therefore, .

Answer:

(i) It is given that D is the midpoint of BC and.

Therefore, …......(1)

Using Pythagoras theorem in the right angled triangle AED,

…......(2)

Let us substitute , and in equation (2), we get

Let us take another right angled triangle that is triangle AEC.

Using Pythagoras theorem,

…......(3)

Let us substitute and in equation (3) we get,

 Here we know that and .

Substituting , and we get

…......(4)

From equation (1) we can substitute in equation (4),

…......(5)

(ii) Using Pythagoras theorem in right angled triangle AEB,

…...... (6)

We know that AB = c and AE = h now we will find BE.

Therefore,

We know that and substituting these values in we get,

Now we will substitute AB = c, AE = h and in equation (6) we get,

…......(7)

Let us rewrite the equation (7) as below,

…......(8)

From equation (1) we can substitute in equation (8),

…......(9)

(iii) Now we will add equations (5) and (9) as shown below,

Therefore, .

Answer:

∆ABC is a right-angled triangle with ∠C = 90°. D is the mid-point of BC.

We need to prove that.

Join AD.

Since D is the midpoint of the side BC, we get

BD = DC

∴

Using Pythagoras theorem in triangles right angled triangle ABC

        .....(1)

Again using Pythagoras theorem in the right angled triangle ADC

       .....(2)

From (1) and (2), we get

Hence, .

Answer:

In ∆ABC and ∆DBA,

    (90º each)

    (Common)

Therefore, by AA-criterion for similarity, we have .

Now we will substitute the values of AC and AB

We are finding the value of AD therefore; we will use the following ratios,

Now we will multiple both sides of the equation by .

We will simplify the above equation as below,

  ..…(1)

But we know that substituting the value of BC in equation (1) we get,

Therefore, the value of AD in terms of b and c is.

Answer:

We are given the following figure with the related information

In the above figure complete the triangle ABC with right angled at C

So

AC = 2 + 2 + 2 + 2 = 8 and

BC = 1 + 1.6 + 1.6 + 1.8 = 6

Using Pythagoras theorem for triangle ABC to find

Hence the distance between A and B is .

Answer:

In order to prove angle it is enough to prove that .

Given,

   .....(1)

Since, so applying Pythagoras theorem in the right angled triangle ABC, we get

      .....(2)

From (1) and (2), we get

Therefore, angle .              (Converse of pythagoras theorem)

Answer:

Given: In ray AD bisects angle A and intersects BC in D, If and

To Prove:

(i)

(ii)

(i) The corresponding figure is as follows

Proof: In triangle ABC, AD is the bisector of

Therefore

Substitute and we get,

By cross multiplication we get.

We proved that

(ii) Since BC = CD + BD

Answer:

In ΔABC in which ∠A is an acute angle with 60°.

Now apply Pythagoras theorem in triangle BCD

Hence

Answer:

Let us draw the figure first.

An aeroplane which flies due north at a speed of 1000 km/hr covers the distance AB after hr and another aeorplane that flies due west at the speed of 1200 km/hr covers the distance BC after hr.

We know that

Let us calculate AB first as shown below,

Similarly we can calculate BC.

Now we have find AC. To find AC we will use Pythagoras theorem,

Taking square root we get,

Therefore, after hrs the aeroplanes will be approximately far apart.

Answer:

We are given ABC is an equilateral triangle with

We have to prove

Draw

In and we have

So by right side criterion of similarity we have

Thus we have

and

Since, therefore

We know that AB = BC = AC

We know that

Substitute in we get

Hence we have proved that

Answer:

Since is right triangle right angled at D

In right , we have

Now substitute

Answer:

Since is right triangle right angled at D

Substitute

Now, in , we have

Therefore,

Hence proved.

Answer:

Given: A rectangle ABCD where AM ⊥ BD and CN ⊥ BD.

To prove: BM² + BN² = DM² + DN²

Proof:

Apply Pythagoras Theorem in ΔAMB and ΔCND,

AB² = AM² + MB²

CD² = CN² + ND²

Since AB = CD, AM² + MB² = CN² + ND²

⇒ AM² − CN² = ND² − MB² … (i)

Again apply Pythagoras Theorem in ΔAMD and ΔCNB,

AD² = AM² + MD²

CB² = CN² + NB²

Since AD = BC, AM² + MD² = CN² + NB²

⇒ AM² − CN² = NB² − MD² … (ii)

Equating (i) and (ii),

ND² − MB² = NB² − MD²

I.e., BM² + BN² = DM² + DN²

This proves the given relation.

Answer:

We have the following figure.

Here is a right triangle right angled at D. Therefore by Pythagoras theorem we have

Again is a right triangle right angled at D.

Therefore, by Pythagoras theorem, we have

Since angle ABD is 45°and therefore angle BAD is also 45°.

Hence AB = DB

So,

Since

So,

Hence we have proved that

Answer:

Let us take

AB = Tip of fishing rod above the surface of the water

BC = The string rest from directly under the tip of the rod.

AC = The length of string

In ABC right triangle right angled at B

Hence 3 m string is out

The string pulled in 12 seconds is equal to to point D 

In this case the diagram will look like the following

Now The length of the new string AD = AC − CD = 3.00 − 0.6 = 2.4 m

Now in triangle ADB we have

∴ Required distance =

Hence, the horizontal distance is .

Answer:

TO STATE: The basic proportionality theorem and its converse.

BASIC PROPORTIONALITY THEOREM: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

CONVERSE OF BASIC PROPORTIONALITY THEOREM: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Answer:

GIVEN: In the figure we are given AD = 6cm, BD = 9cm, AE = 8cm

TO FIND: AC

According to BASIC PROPORTIONALITY THEOREM, if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In ∆ABC, DE || BC. So,

Now,

Answer:

GIVEN: AB = 6cm, BD = 3cm and DC = 2cm. Also, AD is the bisector of .

TO FIND: AC

SOLUTION: We know that the internal bisector of a triangle divides the opposite side internally in the ratio of the sides containing the angle. Therefore,

Answer:

Given: $△\mathrm{ABC}~△\mathrm{PQR}$
Also, $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{1}{3}$
We know if two triangles are similar then the ratio of the areas of two similar triangles is equal to the square of the ratio of their 
corresponding sides. 
$\frac{\mathrm{ar}△\mathrm{ABC}}{\mathrm{ar}△\mathrm{PQR}}={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^2={\left(\frac{1}{3}\right)}^2=\frac{1}{9}$
 

Answer:

SSS Similarity Criterion: If the corresponding sides of two triangles are proportional, then they are similar.

In ΔABC and ΔDEF, if

Then,

Answer:

SAS Similarity Criterion: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then two triangles are similar.

In ΔABC and ΔDEF, if

Then,

Answer:

GIVEN: DE is parallel to BC, AD = 1cm and BD = 2cm.

TO FIND: Ratio of ΔABC to area of ΔADE

According to BASIC PROPORTIONALITY THEOREM, if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In ΔABC, DE || BC.

So

Answer:

GIVEN: AD = 2.4cm, BD = 3.6cm and AC = 5cm.

TO FIND: AE

According to BASIC PROPORTIONALITY THEOREM If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In ∆ABC, DE || BC.

Answer:

Given:  ΔABC and ΔPQR are similar triangles. Area of ΔABC: Area of ΔPQR = 9:16 and BC = 4.5cm.

To find: Length of QR

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Hence,

Answer:

Let ∆ABC and ΔPQR are similar triangles. The area of triangles is 169cm² and 121cm², respectively.

Longest side of the larger triangle is 26cm

TO FIND: length of longest side of the smaller side.

Suppose longest side of the larger triangle is BC and longest side of the smaller triangle is QR.

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Hence,

Answer:

GIVEN: There are two similar triangles ΔABC and ΔDEF.

^,

TO FIND: measure of

SAS Similarity Criterion: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then two triangles are similar.

In ΔABC and ΔDEF if

Then,

So,

Similarly

Now we know that sum of all angles of a triangle is equal to 180°,

Answer:

GIVEN: Altitudes of two similar triangles are in ratio 2:3.

TO FIND: Ratio of the areas of two similar triangles.

Let first triangle be ΔABC and the second triangle be ΔPQR

We know that the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.

Answer:

GIVEN: ΔABC and ΔDEF are two triangles such that .

TO FIND:

We know that two triangles are similar if their corresponding sides are proportional.

Here, ΔABC and ΔDEF are similar triangles because their corresponding sides are given proportional, i.e.

Since the ratio of the areas of two similar triangle is equal to the ratio of the squares of their corresponding sides.

Answer:

GIVEN: ΔABC and ΔDEF are similar triangles such that AB = 3cm, BC = 2cm, CA = 2.5cm and EF = 4cm.

TO FIND: Perimeter of ΔDEF.

We know that if two triangles are similar then their corresponding sides are proportional.

Hence,

Substituting the values, we get

Similarly,

Answer:

TO STATE: Pythagoras theorem and its converse.

PYTHAGORAS THEOREM: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

CONVERSE OF PYTHAGORAS THEOREM: In a triangle, if the square of one side is equal to the sum of the square of the other two sides then the angle opposite to the greatest side is a right angle.

Answer:

GIVEN: the lengths of the diagonals of a rhombus are 30 cm and 40 cm.

TO FIND: side of the rhombus.

Let the diagonals AC and CD of the rhombus ABCD meet at point O.

We know that the diagonals of the rhombus bisect each other perpendicularly.

Hence in right triangle AOD, by Pythagoras theorem

Hence the side of the rhombus is

Answer:

GIVEN: In the given figure PQ || BC, and AP: PB = 1:2

TO FIND:

We know that according to basic proportionality theorem if a line is drawn parallel to one side of a triangle intersecting the other side, then it divides the two sides in the same ratio.

Since triangle APQ and ABC are similar

Hence,

Now, it is given that .

So;

Since the ratio of the areas of two similar triangle is equal to the ratio of the squares of their corresponding sides.

$\frac{\mathrm{Area}\left(\mathrm{APQ}\right)}{\mathrm{Area}\left(\mathrm{ABC}\right)}={\left(\frac{\mathrm{AP}}{\mathrm{AB}}\right)}^2={\left(\frac{1}{3}\right)}^2=\frac{1}{9}$

Hence we got the result

Answer:

Given: In ΔPQR, S and T are the points on the sides PQ and PR respectively such that PT = 2cm, TR = 4cm and ST is parallel to QR.

To find: Ratio of areas of ΔPST and ΔPQR

Now, we know that the areas of two similar triangles are in the ratio of the squares of the corresponding sides. Therefore,

Answer:

Given:

AK = 10 cm

BC = 3.5 cm

HK = 7 cm

To find: AC

Since, so their corresponding sides are proportional.

Answer:

Given: In ∆ABC, DE || BC. BC = 8 cm, AB = 6 cm and DA = 1.5 cm.

To find: DE

In ∆ABC and ∆ADE

$$\begin{gathered} \angle \mathrm{B}=\angle \mathrm{ADE} \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle \mathrm{A}=\angle \mathrm{A} \left(\mathrm{Common}\right) \\ \therefore ∆\mathrm{ABC}~∆\mathrm{ADE} \left(\mathrm{AA} \mathrm{Similarity}\right) \end{gathered}$$

So,

Answer:

Given: In ∆ABC, DE || BC. and BC = 4.5 cm.

To find: DE

Answer:

In ∆ABC and ∆ADE,

∠B = ∠ADE          (Corresponding angles)

∠A = ∠A               (Common)

∴ ∆ABC $~$ ∆ADE     (AA Similarity)

$\Rightarrow \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{ADE}\right)}={\left(\frac{\mathrm{AB}}{\mathrm{AD}}\right)}^2$           (The ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides)

$\Rightarrow \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{ADE}\right)}={\left(\frac{3 \mathrm{cm}}{1 \mathrm{cm}}\right)}^2$         (AB = AD + BD = 1 cm + 2 cm = 3 cm)

$\Rightarrow \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{ADE}\right)}=\frac{9}{1}$

∴ ar(ΔABC) : (ΔADE) = 9 : 1

Thus, the ratio of the area(ΔABC) to the area(ΔADE) is 9 : 1.

Answer:

In ∆ABC, DE || BC.

Using basic proportionality theorem, we have

$$\begin{gathered} \frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{CE}} \\ \Rightarrow \frac{\mathrm{AD}}{7.2 \mathrm{cm}}=\frac{1.8 \mathrm{cm}}{5.4 \mathrm{cm}} \\ \Rightarrow \mathrm{AD}=\frac{1.8 \times 7.2}{5.4}=2.4 \mathrm{cm} \end{gathered}$$

Thus, the length of the side AD is 2.4 cm.

Answer:

∆ABC is an isosceles triangle right angled at C.

∴ BC = AC       (In an isosceles triangle, two sides are equal)

⇒ BC = 4 cm     (AC = 4 cm)

In right ∆ABC,

$$\begin{gathered} {\mathrm{AC}}^2+{\mathrm{BC}}^2={\mathrm{AB}}^2 \left(\mathrm{Pythagoras} \mathrm{theorem}\right) \\ \Rightarrow {\mathrm{AB}}^2={\left(4\right)}^2+{\left(4\right)}^2 \\ \Rightarrow {\mathrm{AB}}^2={\left(4\right)}^2\times 2 \\ \Rightarrow \mathrm{AB}=4\sqrt{2} \mathrm{cm} \end{gathered}$$

Thus, the length of AB is $4\sqrt{2} \mathrm{cm}$.

Answer:

In ∆ABC and ∆PQR,

$\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}}$      (Given)

∴ ∆PQR $~$ ∆CAB     (SSS Similarity criterion)

Here, P ↔ C, Q ↔ A and R ↔ B

If in two triangles ABC and PQR, $\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}}, \mathrm{then} ∆\mathrm{PQR} ~\underline{ ∆\mathrm{CAB} }$.

Answer:

In ΔABC, D and E are points on the sides AB and AC, respectively such that DE || BC.



In ∆ADE and ∆ABC,

∠DAE = ∠BAC     (Common)

∠AED = ∠ACB     (Corresponding angles)

∴ ∆ADE $~$ ∆ABC     (AA Similarity)

$\Rightarrow \frac{\mathrm{DE}}{\mathrm{BC}}=\frac{\mathrm{AD}}{\mathrm{AB}}$        (If two triangles are similar, then their corresponding sides are proportional)

$$\begin{gathered} \Rightarrow \frac{\mathrm{DE}}{7.5 \mathrm{cm}}=\frac{2 \mathrm{cm}}{5 \mathrm{cm}} \left(\mathrm{AB}=\mathrm{AD}+\mathrm{BD}\right) \\ \Rightarrow \mathrm{DE}=\frac{7.5 \times 2}{5}=3 \mathrm{cm} \end{gathered}$$

D and E are respectively the points on the sides AB and AC of ΔABC such that AD = 2 cm, BD = 3 cm, BC =  7.5 cm and DE || BC. Then DE = _____3 cm______.

Answer:

In ∆DEF and ∆PQR,

∠D = ∠Q       (Given)

∠E = ∠R        (Given)

∴ ∆DEF $~$ ∆QRP       (AA Similarity criterion)

$\Rightarrow \frac{\mathrm{DE}}{\mathrm{QR}}=\frac{\mathrm{DF}}{\mathrm{QP}}=\frac{\mathrm{EF}}{\mathrm{RP}}$    (If two triangles are similar, then their corresponding sides are proportional)

If in two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E, then $\frac{\mathrm{DE}}{\mathrm{QR}}=\frac{\mathrm{DF}}{\mathrm{PQ}}=\underline{ \frac{\mathrm{EF}}{\mathrm{RP}} }$.

Answer:

If ∆ABC $~$∆EDF, then $\underline{ \angle \mathrm{A}=\angle \mathrm{E}, \angle \mathrm{B}=\angle \mathrm{D}, \angle \mathrm{C}=\angle \mathrm{F} \mathrm{and} \frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{BC}}{\mathrm{DF}}=\frac{\mathrm{CA}}{\mathrm{FE}} }$.

Answer:

In ΔABC and ΔDEF,

∠B = ∠E    (Given)

∠C = ∠F     (Given)

∴ ΔABC $~$ ΔDEF    (AA Similarity)

Since AB = 3DE, so the two corresponding sides of the given triangles are not equal. Thus, ΔABC and ΔDEF are not congruent.

In ΔABC and ΔDEF, if ∠B = ∠E, ∠F = ∠C and AB = 3DE. Then, the two triangles are _similar_ but not _congruent_.

Answer:

ΔABC ~ ΔQRP      (Given)

$\therefore \frac{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}={\left(\frac{\mathrm{BC}}{\mathrm{RP}}\right)}^2$      (The ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides)

$$\begin{gathered} \Rightarrow {\left(\frac{15 \mathrm{cm}}{\mathrm{RP}}\right)}^2=\frac{9}{4} \\ \Rightarrow \frac{15 \mathrm{cm}}{\mathrm{RP}}=\frac{3}{2} \\ \Rightarrow \mathrm{RP}=\frac{15 \times 2}{3}=10 \mathrm{cm} \end{gathered}$$

If ΔABC ~ ΔQRP, $\frac{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}=\frac{9}{4},$ AB = 18 cm and BC = 15 cm, then PR = ____10 cm____.

Answer:

ΔABC is an equilateral triangle. D and E are mid-points on the sides AB and AC, respectively.



Using mid-point theorem, we have

DE || BC and $\mathrm{DE}=\frac{1}{2}\mathrm{BC}$      .....(1)     

In ∆ADE and ∆ABC,

∠DAE = ∠BAC     (Common)

∠AED = ∠ACB     (Corresponding angles)

∴ ∆ADE $~$ ∆ABC     (AA Similarity)

$\Rightarrow \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{ADE}\right)}={\left(\frac{\mathrm{BC}}{\mathrm{DE}}\right)}^2$        (The ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides)

$\Rightarrow \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{ADE}\right)}={\left(\frac{2\mathrm{DE}}{\mathrm{DE}}\right)}^2=\frac{4}{1}$     [Using (1)]

Or ar(ΔABC) : ar(ΔADE) = 4 : 1

In an equilateral triangle ABC, D is the mid-point of AB  and E is the mid point of AC. The ar(ΔABC) : ar (ΔADE) = 4 : 1.

Answer:

In ΔABC and ΔDEF, $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{FD}}$.

Two triangles are similar if their corresponding sides are proportional.

∴ ΔABC $~$ ΔDEF when $\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{BC}}{\mathrm{DF}}=\frac{\mathrm{CA}}{\mathrm{FE}}$.

If in ΔABC and ΔDEF, $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{FD}}$, then they will be similar, when $\underline{ \frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{BC}}{\mathrm{DF}}=\frac{\mathrm{CA}}{\mathrm{FE}} }$.

Answer:

ΔABC ~ ΔPQR        (Given)

$\therefore \frac{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}{\mathrm{ar}\left(∆\mathrm{ABC}\right)}={\left(\frac{\mathrm{QR}}{\mathrm{BC}}\right)}^2$   (The ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides)

$\Rightarrow \frac{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}{\mathrm{ar}\left(∆\mathrm{ABC}\right)}={\left(\frac{3}{1}\right)}^2=\frac{9}{1}$                 $\left(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{1}{3}\right)$ 

Or $\frac{\mathrm{ar}\left(\mathrm{\Delta PRQ}\right)}{\mathrm{ar}\left(∆\mathrm{BCA}\right)}=9$

It is given that ΔABC ~ ΔPQR, with $\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{1}{3}, \mathrm{then} \frac{\mathrm{ar}\left(\mathrm{\Delta PRQ}\right)}{\mathrm{ar}\left(∆\mathrm{BCA}\right)}$ = _____9_____.

Answer:

ΔABC ~ ΔDEF      (Given)

∴ ∠C = ∠F      (If two triangles are similar, then their corresponding angles are equal)

⇒ ∠F = 50°     (∠C = 50°)

Also,

$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}$    (If two triangles are similar, then their corresponding sides are proportional)

$$\begin{gathered} \Rightarrow \frac{5 \mathrm{cm}}{\mathrm{DE}}=\frac{8 \mathrm{cm}}{7.5 \mathrm{cm}} \\ \Rightarrow \mathrm{DE}=\frac{5 \times 7.5}{8}=4.7 \mathrm{cm} \left(\mathrm{Approx}\right) \end{gathered}$$

It is given that ΔABC ~ ΔDEF, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm. Then DE = __4.7 cm __ and ∠F = ___50°___.

Answer:

In ΔABC, AB = 24 cm, BC = 10 cm and AC = 26 cm.

Now,

AB² + BC² = (24 cm)² + (10 cm)² = 576 cm² + 100 cm² = 676 cm²

AC² = (26 cm)² = 676 cm²

∴ AB² + BC² = AC²

In ΔABC, AB² + BC² = AC²

Using converse of Pythagoras theorem, we have

ΔABC is a right triangle right angled at B.

⇒ ∠B = 90º   

In ΔABC, if AB = 24 cm, BC = 10 cm and AC = 26 cm, then ∠B = ____90º  ____.

Answer:

Let the length of the side of the rhombus be a cm.

We know

(Length of one diagonal of rhombus)² + (Length of other diagonal of rhombus)² = 4 × (Length of side of rhombus)²

$$\begin{gathered} \therefore 4a^2={\left(16\right)}^2+{\left(12\right)}^2 \\ \Rightarrow 4a^2=256+144 \\ \Rightarrow 4a^2=400 \end{gathered}$$
$$\begin{gathered} \Rightarrow a^2=100 \\ \Rightarrow a=10 \mathrm{cm} \end{gathered}$$

Thus, the length of the side of the rhombus is 10 cm.

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Then the length of the side of the rhombus is ___10 cm___.

Answer:

In an isosceles ∆ABC,

AB² = 2AC²       (Given)

⇒ AB² = AC² + AC²

⇒ AB² = AC² + BC²           (AC = BC)

Using converse of Pythagoras theorem, we have

∆ABC is an isosceles right triangle right angled at C.

(In a triangle, if the square of one side is equal to the sum of the squares of other two sides, then the angle opposite to side is a right angle.)  

∴ ∠C = 90º          

In an isosceles triangle ABC, if AC = BC and AB² = 2AC², then ∠C = ___90º___.

Answer:

In ∆ABD,

∠ABD + ∠BAD = 90º    .....(1)

Now, ∠BAC = 90°

Or ∠CAD + ∠BAD = 90º        .....(2)

From (1) and (2), we have

∠ABD + ∠BAD = ∠CAD + ∠BAD

⇒ ∠ABD = ∠CAD

In ∆ABD and ∆ACD,

∠ADB = ∠ADC     (90º each)

∠ABD = ∠CAD      (Proved above)

∴ ∆ABD $~$ ∆CAD    (AA Similarity)

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{CA}}=\frac{\mathrm{AD}}{\mathrm{CD}}=\frac{\mathrm{BD}}{\mathrm{AD}}$   (If two triangles are similar, then their corresponding sides are proportional)

$$\begin{gathered} \Rightarrow \frac{\mathrm{AD}}{\mathrm{CD}}=\frac{\mathrm{BD}}{\mathrm{AD}} \\ \Rightarrow \mathrm{BD}.\mathrm{CD}={\mathrm{AD}}^2 \end{gathered}$$

In Fig. 7.238, ∠BAC = 90° and AD ⊥ BC. Then, BD. CD = ____AD²____.