Rd Sharma 2022 _mcqs for Class 10 Maths Chapter 7 - Triangles

Answer:

Given: Sides of two similar triangles are in the ratio 4:9

To find: Ratio of area of these triangles

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

Hence the correct answer is option

Answer:

Given: Areas of two similar triangles are 9cm² and 16cm².

To find: Ratio of their corresponding sides.

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

$\frac{\mathrm{side}1}{\mathrm{side}2}=\frac{3}{4}$

So, the ratio of their corresponding sides is 3 : 4.

Hence the correct answer is .

Answer:

Given: Areas of two similar triangles ΔABC and ΔDEF are 144cm² and 81cm².

If the longest side of larger ΔABC is 36cm

To find: the longest side of the smaller triangle ΔDEF

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

$\frac{12}{9}=\frac{36}{\mathrm{longest} \mathrm{side} \mathrm{of} \mathrm{smaller} ∆\mathrm{DEF}}$

= 27 cm

Hence the correct answer is

Answer:

Given: ΔABC and ΔBDE are two equilateral triangles such that D is the midpoint of BC.

To find: Ratio of areas of ΔABC and ΔBDE.



ΔABC and ΔBDE are equilateral triangles; hence they are similar triangles.

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

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

Hence the correct answer is .

Answer:

Given: ΔABC and ΔDEF are similar triangles such that 2AB = DE and BC = 8 cm.

To find: EF

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

Hence, for similar triangles ΔABC and ΔDEF

Hence the correct answer is .

Answer:

Given: ΔABC and ΔDEF are two triangles such that .

To find:

We know that if the sides of two triangles are proportional, then the two triangles are similar.

Since , therefore, ΔABC and ΔDEF are similar.

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

Hence the correct answer is .

Answer:

Given: XY is drawn parallel to the base BC of a ΔABC cutting AB at X and AC at Y. AB = 4BX and YC = 2 cm.

To find: AY

In ΔAXY and ΔABC,

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

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

It is given that AB = 4BX.

Let AB = 4x and BX = x.

Then, AX = 3x

Hence the correct answer is .

Answer:

Given: Two poles of heights 6m and 11m stand vertically upright on a plane ground. Distance between their foot is 12 m.

To find: Distance between their tops.

Let CD be the pole with height 6m.

AB is the pole with height 11m, distance between their foot i.e. DB is 12 m.

Let us assume a point E on the pole AB which is 6m from the base of AB.

Hence

AE = AB − 6 = 11 − 6 = 5 m

Now in right triangle AEC, Applying Pythagoras theorem

AC² = AE² + EC²

AC² = 5² + 12²                  (since CDEB forms a rectangle and opposite sides of rectangle are equal)

AC² = 25 + 144

AC² = 169

Thus, the distance between their tops is 13m.

Hence correct answer is .

Answer:

Given: In ΔABC, D and E are points on the side AB and AC respectively such that DE || BC and AD : DB = 3 : 1. Also, EA = 3.3cm.

To find: AC

In ∆ABC, DE || BC.

Using corollory of basic proportionality theorem, we have

Hence the correct answer is .

Answer:

Given: In ΔABC and ΔDEF

To find: Measure of angle B.

In ΔABC and ΔDEF

Hence in similar triangles ΔABC and ΔDEF

We know that sum of all the angles of a triangle is equal to 180°.

Hence the correct answer is .

Answer:

Given: If ΔABC and ΔDEF are similar triangles such that

To find: Measure of angle C

In similar ΔABC and ΔDEF,

We know that sum of all the angles of a triangle is equal to 180°.

Hence the correct answer is

Answer:

GIVEN: In ΔABC, D, E and F are the midpoints of BC, CA, and AB respectively.

TO FIND: Ratio of the areas of ΔDEF and ΔABC

Since it is given that D and, E are the midpoints of BC, and AC respectively.

Therefore DE || AB, DE || FA……(1)

Again it is given that D and, F are the midpoints of BC, and, AB respectively.

Therefore, DF || CA, DF || AE……(2)

From (1) and (2) we get AFDE is a parallelogram.

Similarly we can prove that BDEF is a parallelogram.

Now, in ΔADE and ΔABC

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

Hence the correct option is .

Answer:

Given: In ΔABC,, AC = 12cm, and AB = 5cm.

To find: AD

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

In ∆ACB and ∆ADC,

$\angle \mathrm{C}=\angle \mathrm{C}$        (Common)

$\angle \mathrm{A}=\angle \mathrm{ADC}=90°$  

∴ ∆ACB $~$∆ADC     (AA Similarity)

We got the result as

Answer:

Given: In an equilateral ΔABC, .

Since , BD = CD = $\frac{\mathrm{BC}}{2}$

Applying Pythagoras theorem,

In ΔADC

We got the result as

Answer:

Given: In a ΔABC, AD is the bisector of . AB = 6cm and AC = 5cm and BD = 3cm.

To find: DC

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

Hence,

Hence we got the result

Answer:

Given: In a ΔABC, AD is the bisector of angle BAC. AB = 8cm, and DC = 3cm and BD = 6cm.

To find: AC

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

Hence,

Hence we got the result

Answer:

Given: ABCD is a trapezium in which BC||AD and AD = 4 cm

The diagonals AC and BD intersect at O such that

To find: DC

 In ΔAOD and ΔCOB

$$\begin{gathered} \angle \mathrm{OAD}=\angle \mathrm{OCB} \left(\mathrm{Alternate} \mathrm{angles}\right) \\ \angle \mathrm{ODA}=\angle \mathrm{OBC} \left(\mathrm{Alternate} \mathrm{angles}\right) \\ \angle \mathrm{AOD}=\angle \mathrm{BOC} \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \mathrm{So}, ∆\mathrm{AOD}~∆\mathrm{COB} \left(\mathrm{AAA} \mathrm{similarity}\right) \\ \mathrm{Now}, \mathrm{correponding} \mathrm{sides} \mathrm{of} \mathrm{similar} ∆\text{'}\mathrm{s} \mathrm{are} \mathrm{proportional}. \\ \frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{AD}}{\mathrm{BC}} \\ \Rightarrow \frac{1}{2}=\frac{\mathrm{AD}}{\mathrm{BC}} \\ \Rightarrow \frac{1}{2}=\frac{4}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=8 \mathrm{cm} \end{gathered}$$

Hence the correct answer is

Answer:

M is the mid-point of AB.

∴ $\mathrm{BM}=\frac{\mathrm{AB}}{2}$

N is the mid-point of BC.

∴ $\mathrm{BN}=\frac{\mathrm{BC}}{2}$

Now,

Hence option (b) is correct.

Answer:

Given: In ΔABC and ΔDEF, .

We know that if in two triangles, one pair of corresponding sides are proportional and the included angles are equal, then the two triangles are similar.

Hence, ΔABC is similar to ΔDEF, we should have.

Hence the correct answer is .

Answer:

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

∴ ΔCAB $~$ΔFDE

Hence the correct answer is .

Answer:

Given:

To find: measure of EF

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

Hence the correct answer is

Answer:

Given: One side of isosceles right triangle is 4√2cm

To find: Length of the hypotenuse.

We know that in isosceles triangle two sides are equal.

In isosceles right triangle ABC, let AB and AC be the two equal sides of measure 4√2cm.

Applying Pythagoras theorem, we get

Hence correct answer is .

Answer:

A man goes 24m due to west and then 7m due north.

Let the man starts from point B and goes 24 m due to west and reaches point A, then walked 7m north and reaches point C.

Now we have to find the distance between the starting point and the end point i.e. BC.

In right triangle ABC, applying Pythagoras theorem, we get

Hence correct answer is .

Answer:

Given: In Δ ABC and Δ DEF

To find: Ar(Δ DEF)

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

Hence the correct answer is

Answer:

Given: In Δ ABC and ΔPQR

To find: Measure of QR

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

Hence the correct answer is

Answer:

Given: The area of two similar triangles is 121cm² and 64cm² respectively. The median of the first triangle is 2.1cm.

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.

Taking square root on both side, we get

$$\begin{gathered} \frac{11}{8}=\frac{12.1\mathrm{cm}}{\mathrm{median}2} \\ \Rightarrow \mathrm{median}2= 8.8 \mathrm{cm} \end{gathered}$$

Hence the correct answer is .

Answer:

Given that, an isosceles right triangle has a hypotenuse measuring 10 cm. Let the triangle be $△$ABC, right-angled at B.

Using Pythagoras theorem,
$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2 \\ \Rightarrow {\left(10\right)}^2={\mathrm{AB}}^2+{\mathrm{AB}}^2 \\ \Rightarrow 100=2{\mathrm{AB}}^2 \\ \Rightarrow {\mathrm{AB}}^2=50 \\ \Rightarrow \mathrm{AB}=\mathrm{BC}=5\sqrt{2} \mathrm{cm} \left(\because \mathrm{side} \mathrm{cannot} \mathrm{be} \mathrm{negative}\right) \end{gathered}$$

Therefore,
$\begin{array}{rcl}\mathrm{Perimeter} \mathrm{of} △\mathrm{ABC}& =& 10+5\sqrt{2}+5\sqrt{2}\\ & =& 10+10\sqrt{2}\\ & =& 10\left(1+\sqrt{2}\right) \mathrm{cm}\end{array}$

Hence, the correct answer is option (c).

Answer:

∆ABC is an equilateral triangle and .

In ∆ABD, applying Pythagoras theorem, we get

We got the result as .

Answer:

It is given that in $∆\mathrm{ABC}$, AC = BC
Also, it is given that AB² = 2AC^{2
$$\begin{gathered} \Rightarrow {\mathrm{AB}}^2={\mathrm{AC}}^2+{\mathrm{AC}}^2 \\ \Rightarrow {\mathrm{AB}}^2={\mathrm{AC}}^2+{\mathrm{BC}}^2 (\mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{AC}=\mathrm{BC}) \end{gathered}$$}
This is a condition for the Pythagoras theorem.
Therefore, $∆\mathrm{ABC}$ is a right angled triangle, where AB is the hypotenuse and AC and BC are the other sides. 
$\Rightarrow \angle \mathrm{C}={90}^{\circ }$
Hence, correct answer is option (c).

Answer:

Given: In an isosceles ΔABC, , AC = 6 cm.

To find: AB

In an isosceles ΔABC,.

Therefore, BC = AC = 6 cm

Applying Pythagoras theorem in ΔABC, we get

We got the result as .

Answer:

 In ΔABC and ΔDEF

Hence

Hence the correct answer is (b).

Answer:

ΔABC and ΔDEF,

$$\begin{gathered} \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{EF}}{\mathrm{ED}} \\ \angle \mathrm{A}=\angle \mathrm{E}=130° \end{gathered}$$

ΔABC $~$ ΔEFD   (SAS Similarity)

$$\begin{gathered} \therefore \angle \mathrm{F}=\angle \mathrm{B}=30° \\ \angle \mathrm{D}=\angle \mathrm{C}=20° \end{gathered}$$

Hence the correct answer is

Answer:

Given: In ∆ABC,  DE || AB.

To find: the value of x

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 || AB

                          x = 2

Hence we got the result .

Answer:

Given:

To find: The value of CE

Since   

∴ DE || BC       (Two lines are parallel if the corresponding angles formed are equal)

According to basic proportionality theorem if a line is 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

Hence we got the result .

Answer:

Given: RS || DB || PQ. CP = PD = 11cm and DR = RA = 3cm

To find: the value of x and y respectively.

$$\begin{gathered} \mathrm{In} ∆\mathrm{ASR} \mathrm{and} ∆\mathrm{ABD}, \\ \angle \mathrm{ASR}=\angle \mathrm{ABQ} \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle \mathrm{A}=\angle \mathrm{A} \left(\mathrm{Common}\right) \\ \therefore ∆\mathrm{ASR} ~∆\mathrm{ABD} \left(\mathrm{AA}\hspace{0.17em}\mathrm{Similarity}\right) \end{gathered}$$

This relation is satisfied by option (d).

Hence, x = 16 cm and y = 8cm

Hence the result is .

Answer:

Given: PB||CF and DP||EF. AB = 2 cm and AC = 8 cm.

To find: AD: DE

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 ∆ACF, PB || CF.

  .....(1)

Again, DP||EF.

Hence we got the result .

Answer:

In right ∆OAB,

$$\begin{gathered} {\mathrm{AB}}^2={\mathrm{OA}}^2+{\mathrm{OB}}^2 \left(\mathrm{Pythagoras} \mathrm{Theorem}\right) \\ \Rightarrow {\mathrm{AB}}^2={\left(10\right)}^2+{\left(10\right)}^2 \left(\mathrm{OA}=\mathrm{OB}=10 \mathrm{cm}\right) \\ \Rightarrow {\mathrm{AB}}^2=100+100=200 \\ \Rightarrow \mathrm{AB}=\sqrt{200}=10\sqrt{2} \mathrm{cm} \end{gathered}$$

Thus, the length of the chord is $10\sqrt{2}$ cm.

Hence, the correct answer is option B.

Answer:

Given: Vertical stick 20m long casts a shadow 10m long on the ground. At the same time a tower casts the shadow 50 m long on the ground.

To determine: Height of the tower

Let AB be the vertical stick and AC be its shadow. Also, let DE be the vertical tower and DF be its shadow.

Join BC and EF.

In ΔABC and ΔDEF, we have

We know that in any two similar triangles, the corresponding sides are proportional. Hence,

Hence the correct answer is option .

Answer:

Given: Two isosceles triangles have equal vertical angles and their areas are in the ratio of 16:25.

To find: Ratio of their corresponding heights.

Let ∆ABC and ∆PQR be two isosceles triangles such that $\angle \mathrm{A}=\angle \mathrm{P}$. Suppose AD ⊥ BC and PS ⊥ QR .

In ∆ABC and ∆PQR,

$$\begin{gathered} \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}} \\ \angle \mathrm{A}=\angle \mathrm{P} \\ \therefore ∆\mathrm{ABC}~∆\mathrm{PQR} \left(\mathrm{SAS} \mathrm{similarity}\right) \end{gathered}$$

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

Hence,

$$\begin{gathered} \frac{\mathrm{Ar}\left(∆\mathrm{ABC}\right)}{\mathrm{Ar}\left(∆\mathrm{PQR}\right)}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^2 \\ \Rightarrow \frac{16}{25}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^2 \\ \Rightarrow \frac{\mathrm{AD}}{\mathrm{PS}}=\frac{4}{5} \end{gathered}$$

Hence we got the result as

Answer:

Given: In ΔABC, 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 sides are proportional

Since ΔABC and ΔDEF are similar,

From (1) and (2), we get

Perimeter of ΔDEF =  DE + EF + FD = 6 + 4 +5 = 15 cm

Hence the correct answer is .

Answer:

Given: XY||BC and BY is bisector of $\angle$XYC.

Since XY||BC

So $\angle$YBC = $\angle$BYC     (Alternate angles)

Now, in triangle BYC two angles are equal. Therefore, the two corresponding sides will be equal.

Hence, BC = CY

Hence option (a) is correct.

Answer:

Given: In ΔABC,, AC = 12cm, and AB = 5cm.

To find: AD

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

In ∆ACB and ∆ADC,

$\angle \mathrm{C}=\angle \mathrm{C}$        (Common)

$\angle \mathrm{A}=\angle \mathrm{ADC}=90°$  

∴ ∆ACB $~$∆ADC     (AA Similarity)

We got the result as .

Answer:

Given: In ΔABC,, BD = 8cm, DC = 2 cm and AD = 4cm.

In ΔADC,

Similarly, in ΔADB

Now, In ΔABC

${\mathrm{BC}}^2={\left(\mathrm{CD}+\mathrm{DB}\right)}^2={\left(2+8\right)}^2={\left(10\right)}^2=100$

and

${\mathrm{AB}}^2+{\mathrm{CA}}^2=80+20=100$

$\therefore {\mathrm{AB}}^2+{\mathrm{CA}}^2={\mathrm{BC}}^2$

Hence, triangle ABC is right angled at A.

We got the result as

Answer:

Given: In ΔABC, D is on side AB and point E is on side AC, such that BCED is a trapezium. DE: BC = 3:5.

To find: Calculate the ratio of the areas of ΔADE and the trapezium BCED.

In ΔADE and ΔABC,

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

We know that

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

Now ,

Hence the correct answer is.

Answer:

Given: ΔABC is an isosceles triangle, D is a point on BC such that

We know that in an isosceles triangle the perpendicular from the vertex bisects the base.

∴ BD = DC

Applying Pythagoras theorem in ΔABD

Since

Hence correct answer is .

Answer:

Given: In ΔABC, and.

To find: BD: DC

$$\begin{gathered} \angle \mathrm{CAD}+\angle \mathrm{BAD}=90° .....\left(1\right) \\ \angle \mathrm{BAD}+\angle \mathrm{ABD}=90° .....\left(2\right) \left(\angle \mathrm{ADB}=90°\right) \\ \mathrm{From} \left(1\right) \mathrm{and} \left(2\right), \\ \angle \mathrm{CAD}=\angle \mathrm{ABD} \end{gathered}$$

In ΔADB and ΔADC,

$$\begin{gathered} \angle \mathrm{ADB}=\angle \mathrm{ADC} \left(90° \mathrm{each}\right) \\ \angle \mathrm{ABD}=\angle \mathrm{CAD} \left(\mathrm{Proved}\right) \\ \therefore ∆\mathrm{ADB}~∆\mathrm{ADC} \left(\mathrm{AA} \mathrm{Similarity}\right) \\ \Rightarrow \frac{\mathrm{CD}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\mathrm{AD}}{\mathrm{BD}} \left(\mathrm{Corresponding} \mathrm{sides} \mathrm{are} \mathrm{proportional}\right) \end{gathered}$$

Answer:

In triangle ABC, E is a point on AC such that .

We need to find .



Since , CE = AE = $\frac{\mathit{A}\mathit{C}}{\mathit{2}}$              (In a equilateral triangle, the perpendicular from the vertex bisects the base.)

In triangle ABE, we have

Since AB = BC = AC

Therefore,

Since in triangle BE is an altitude, so

Hence option (c) is correct.

Answer:

Disclaimer: There is mistake in the problem. The question should be "In a right triangle ABC right-angled at B, if P and Q are points on the sides AB and BC respectively, then"

Given: In the right ΔABC, right angled at B. P and Q are points on the sides AB and BC respectively.

Applying Pythagoras theorem,

In ΔAQB,

${\mathrm{AQ}}^2={\mathrm{AB}}^2+{\mathrm{BQ}}^2$       .....(1)

In ΔPBC

       .....(2)

Adding (1) and (2), we get

${\mathrm{AQ}}^2+{\mathrm{CP}}^2={\mathrm{AB}}^2+{\mathrm{BQ}}^2+{\mathrm{PB}}^2+{\mathrm{BC}}^2$    .....(3)

In ΔABC,

      .....(4)

In ΔPBQ,

${\mathrm{PQ}}^2={\mathrm{PB}}^2+{\mathrm{BQ}}^2$        .....(5)

From (3), (4) and (5), we get

${\mathrm{AQ}}^2+{\mathrm{CP}}^2={\mathrm{AC}}^2+{\mathrm{PQ}}^2$

Answer:

Given: ΔABC and ΔDEF are similar triangles such that DE = 3cm, EF = 2cm, DF = 2.5cm and BC = 4cm.

To find: Perimeter of ΔABC.

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

Hence,

Substituting the values we get

Similarly,

Hence the correct option is

Answer:

Given: ΔABC is similar to ΔDEF such that AB= 9.1cm, DE = 6.5cm. Perimeter of ΔDEF is 25cm.

To find: Perimeter of ΔABC.

We know that the ratio of corresponding sides of similar triangles is equal to the ratio of their perimeters.

Hence,

Hence the correct answer is .

Answer:

Given: In Isosceles ΔABC, AC = BC and AB² = 2AC².

To find: Measure of angle C

In Isosceles ΔABC,

AC = BC

$\angle \mathrm{B}=\angle \mathrm{A}$    (Equal sides have equal angles opposite to them)

Hence the correct answer is .

Answer:

Given that, in $△$ABC, if AB = 4 cm, BC = 8 cm and AC = $4\sqrt{3}$ cm.

Using Pythagoras theorem,
$$\begin{gathered} B{C}^2=A{B}^2+A{C}^2 \\ \Rightarrow {\left(8\right)}^2={\left(4\right)}^2+{\left(4\sqrt{3}\right)}^2 \\ \Rightarrow 64=16+48 \\ \Rightarrow 64=64 \end{gathered}$$

Thus, $△$ABC is a right-angle triangle with $\angle A=90°$.

Hence, the correct answer is option (d).

Answer:

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

$$\begin{gathered} \frac{ar(△ABC)}{ar(△PQR)}={\left(\frac{AB}{PQ}\right)}^2 \\ \Rightarrow \frac{ar(△ABC)}{ar(△PQR)}={\left(\frac{1.2}{1.4}\right)}^2 \\ \Rightarrow \frac{ar(△ABC)}{ar(△PQR)}={\left(\frac{6}{7}\right)}^2 \\ \Rightarrow \frac{ar(△ABC)}{ar(△PQR)}=\frac{36}{49} \end{gathered}$$

Hence, the correct answer is option (c).

Answer:

Given: In DPQR, ÐQ = 90°, PQ = 5 cm, QR = 12 cm and QS ^ PR.


Now, in DPQR using Pythagoras theorem, we have

(PR)² = (PQ)² + (QR)²

⇒ (PR)² = (5)² + (12)²

⇒ (PR)² = 25 + 144

⇒ (PR)² = 169

⇒ PR = 13 cm

Area of DPQR = $\frac{1}{2}\times PQ\times QR=\frac{1}{2}\times PR\times QS$

⇒ PQ × QR = PR × QS

$$\begin{gathered} \Rightarrow QS=\frac{PQ\times QR}{PR} \\ \Rightarrow QS=\frac{12\times 5}{13} \\ \Rightarrow QS=\frac{60}{13} \mathrm{cm} \end{gathered}$$

Hence, the correct answer is option (c).



 

Answer:

Given: D is a point on side BC of DABC such that BD = CD = AD



In △ABD,
BD = AD
⇒ ∠ABD = ∠BAD = x (Let)              (∵ angles opposite to equal sides are equal)

Now, In △ADC, 
CD = AD
⇒ ∠DCA = ∠CAD = x (Let)              (∵ angles opposite to equal sides are equal)



Now, in △ABC, using the angle sum property, we have
∠ABC + ∠BCA + ∠BAC = 180° 
⇒ ∠ABD + ∠BAD + ∠CAD + ∠BAC = 180° 
⇒ x + x + x + x = 180° 
⇒ 4x = 180° 
⇒ x = 45° 

Now, ∠BAC = ∠BAD + ∠CAD
⇒ ∠BAC = 45° + 45° = 90°

Thus, △ABC is right angled at ∠​A.

So, using Pythagoras theorem in △ABC, we get
AB² + AC² = BC²

Hence, the correct answer is option (c).

Answer:

Given that, ABCD is a trapezium in which AB || DC and AB = 2 DC. The diagonals AC and BD intersect at O.



Now, in △AOB and △COD, we have
∠AOB = ∠COD                      (Vertically opposite angles)
∠OAB = ∠OCD                      (Alternate interior angles)
∴ △AOB ~ △COD                 (By AA similarity)

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$$\begin{gathered} \therefore \frac{ar\left(AOB\right)}{ar\left(COD\right)}={\left(\frac{AB}{CD}\right)}^2 \\ \Rightarrow \frac{ar\left(AOB\right)}{ar\left(COD\right)}={\left(\frac{2CD}{CD}\right)}^2 \\ \Rightarrow \frac{ar\left(AOB\right)}{ar\left(COD\right)}=4 \\ \Rightarrow ar\left(COD\right)=\frac{ar\left(AOB\right)}{4} \\ \Rightarrow ar\left(COD\right)=\frac{84}{2} \\ \Rightarrow ar\left(COD\right)=21 {\mathrm{cm}}^2 \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

Given: ABC is an isosceles right triangle which is right angled at C and AC = BC.



By using Pythagoras theorem, we get
AC² + BC² = AB²
⇒ AC² + AC² = AB²
⇒ AB² = 2AC²

Hence, the correct answer is option (a).

Answer:

Given that, ∆ABC  is a right triangle right angled at B and BD ⊥ AC.



In △ADB and △ABC,
∠A = ∠A
∠ADB = ∠ADB
So, by AA similarity, △ADB ~ △ABC.

Therefore, all the given options are incorrect.

Hence, the correct answers are options (a), (b), (c) and (d).

Disclaimer: All the options are incorrect as per the calculations given above.

Answer:

In DABC, if AD is the bisector of ∠A

$\therefore \frac{AB}{AC}=\frac{BD}{DC} .....\left(1\right)$

Draw AL ⏊ BC



$$\begin{gathered} \therefore \frac{ar\left(ABD\right)}{ar\left(ACD\right)}=\frac{{\displaystyle \frac{1}{2}}\times BD\times AL}{{\displaystyle \frac{1}{2}}\times CD\times AL} \\ \Rightarrow \frac{ar\left(ABD\right)}{ar\left(ACD\right)}=\frac{{\displaystyle BD}}{{\displaystyle CD}} \\ \Rightarrow \frac{ar\left(ABD\right)}{ar\left(ACD\right)}=\frac{{\displaystyle AB}}{{\displaystyle AC}} \left[\mathrm{From} \left(1\right)\right] \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

Given: O is a point on side PQ of △PQR such that PO = QO = RO



In △POR, 
PO = OR
⇒ ∠OPR = ∠ORP = x (Let)            (∵ angles opposite to equal sides are equal)



Now, in △QOR, 
QO = RO
⇒ ∠OQR = ∠ORQ = x (Let)            (∵ angles opposite to equal sides are equal)

Now, in △PQR, using the angle sum property, we have
∠PQR + ∠PRQ + ∠QPR = 180°  
⇒ x + x + x + x = 180° 
⇒ 4x = 180° 
⇒ x = 45° 

Now, ∠PRQ = ∠ORP + ∠ORQ
⇒ ∠PRQ = 45° + 45° = 90°
Thus, △PQR is right angled at R.

So, using Pythagoras theorem in △PQR, we get
PR² + RQ² = PQ²

Hence, the correct answer is option (b).

Answer:

PQRS is a parallelogram, therefore PS || QR
⇒ AS || QR

Also, PQ = SR, QR = PS 

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

So, in △QTR,
$$\begin{gathered} \frac{AT}{AQ}=\frac{ST}{SR} \\ \Rightarrow \frac{6}{6}=\frac{4}{y} \\ \Rightarrow y=4 \mathrm{cm} \end{gathered}$$

Therefore, y = 4 cm.

Hence, the correct answer is option (d).
 

Answer:

Given that, AP = 3 cm, AR = 4.5 cm, AQ = 6 cm, AB = 5 cm and AC = 10 cm.
Now,
$$\begin{gathered} \frac{AP}{AB}=\frac{3}{5} \\ \frac{AQ}{AC}=\frac{6}{10}=\frac{3}{5} \\ \Rightarrow \frac{AP}{AB}=\frac{AQ}{AC} \end{gathered}$$

According to the converse of Thales theorem, "If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side."
Therefore, PQ ∥ BC.
⇒ PR ∥ BD

In $△$ABD, using Thales theorem,
$$\begin{gathered} \frac{AP}{PB}=\frac{AR}{RD} \\ \Rightarrow \frac{PB}{AP}=\frac{RD}{AR} \\ \Rightarrow \frac{PB}{AP}+1=\frac{RD}{AR}+1 \\ \Rightarrow \frac{AP+PB}{AP}=\frac{AR+RD}{AR} \\ \Rightarrow \frac{AB}{AP}=\frac{AD}{AR} \\ \Rightarrow \frac{5}{3}=\frac{AD}{4.5} \\ \Rightarrow AD=7.5 \mathrm{cm} \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

Consider △PSR and △PQR.
∠PQR = ∠PRS          (given)
∠P = ∠P                     (common)
∴ △PSR ~ △PQR

Thus, the ratio of their corresponding sides is equal.
$$\begin{gathered} \Rightarrow \frac{PQ}{PR}=\frac{PR}{PS} \\ \Rightarrow PQ=\frac{P{R}^2}{PS} \\ \Rightarrow PQ=\frac{8\times 8}{4} \\ \Rightarrow PQ=16 \end{gathered}$$

∴ PQ = 16 cm

Hence, the correct answer is option (b).

Answer:

Given: ∆PQR ~ ∆XYZ 

$$\begin{gathered} \Rightarrow \frac{PQ}{XY}=\frac{QR}{YZ}=\frac{PR}{XZ} \\ \Rightarrow \frac{8}{4}=\frac{QR}{4.5}=\frac{PR}{6.5} \\ \Rightarrow QR=9 \mathrm{cm} \mathrm{and} PR=13 \mathrm{cm} \end{gathered}$$

Therefore,
Perimeter of ∆PQR = 8 + 9 + 13
                                 = 30 cm

Hence, the correct answer is option (d).

Answer:

Statement S-1: ABC is a right triangle provided n² – m² = r².

ABC is a right triangle provided n² – m² = r²
⇒ n² = m² + r²
Thus, n is the hypotenuse, and m and r are base and height of the triangle.

Therefore, statement S-1 is true only if and only if n > m, n > r.

Statement S-2: Triangle with side lengths m + 2, n + 2 and r + 2 is a right angle triangle.
For the right angle triangle, $m^2=n^2+r^2$.

Now,
$$\begin{gathered} {\left(m+2\right)}^2={\left(n+2\right)}^2+{\left(r+2\right)}^2 \\ \Rightarrow m^2+4+4m=n^2+4+4n+r^2+4+4r \\ \Rightarrow 4m=4n+4+4r \\ \Rightarrow m=n+r+1 .....\left(1\right) \end{gathered}$$

$\begin{array}{rcl}{\left(m+2\right)}^2& =& m^2+4+4m\\ & =& {\left(n+r+1\right)}^2+4\left(n+r+1\right)+4 \left\{\mathrm{Using} \left(1\right)\right\}\\ & =& n^2+r^2+1+2nr+2r+2n+4n+4r+4+4\\ & =& n^2+r^2+2nr+6r+6n+9\\ & =& \left(n^2+4n+4\right)+\left(r^2+4r+4\right)+2nr+2n+2r+1\\ & =& {\left(n+2\right)}^2+{\left(r+2\right)}^2+2\left(nr+n+r\right)+1\ne {\left(n+2\right)}^2+{\left(r+2\right)}^2\end{array}$

Therefore, statement S-2 is false.

Statement S-3: Triangle with sides 2m, 2n and 2r is a right-angle triangle.
If triangle ABC with side lengths m, n and r is a right-angle triangle, then the triangle with sides 2m, 2n and 2r is a right-angle triangle.
$\begin{array}{rcl}{\left(2m\right)}^2& =& 4m^2\\ & =& 4\left(n^2+r^2\right) \left\{\because m^2=n^2+r^2\right\}\\ & =& 4n^2+4r^2\\ & =& {\left(2n\right)}^2+{\left(2r\right)}^2\end{array}$

Thus, statement S-3 would be correct if ∆ABC is a right triangle.

Hence, the correct answer is option (c).





 

Answer:

In ∆AOB and ∆DOC,
 $$\begin{gathered} \frac{AO}{DO}=\frac{1.6}{0.8}=2 \\ \frac{BO}{CO}=\frac{4.8}{2.4}=2 \\ \Rightarrow \frac{AO}{DO}=\frac{BO}{CO} \end{gathered}$$

Thus, the corresponding sides of the two triangles have the same ratio.

Also,
∠AOB = ∠DOC                       (Vertically opposite angles)

Thus, ∆AOB ~ ∆DOC by SAS criterion.

Hence, the correct answer is option (d).

Answer:

From the figure, we get that
∠AOB = ∠COD = 60° (Vertically opposite angles)

Statement-1: ∠B = 70°
Using the angle sum property in △AOB, we get
∠A + ∠O + ∠B = 180°
⇒ 7x + 1° + 60° + 70° = 180°
⇒ 7x = 49°
⇒ x = 7°
⇒ ∠A = 7 × 7° + 1° = 50°

Now, in △COD,
∠D = 3x + 29°
       = 3 × 7° + 29°
       = 50°

Using the angle sum property in △COD, we get
∠C + ∠O + ∠D = 180°
​⇒ ∠C + 60° + 50° = 180°
⇒ ∠C = 70°

So, in ∠AOB and ∠COD, we have
∠AOB = ∠COD                     (vertically opposite angles)
∠A = ∠D
∠B = ∠C
Thus, △AOB ~ △DOC.

Statement-2: ∠C = 70°
Using the angle sum property in △COD, we get
∠C + ∠O + ∠D = 180°
​⇒ 70° + 60° + 3x + 29° = 180°
⇒ x = 7°
Thus, ∠D = 3x + 29° = 50°

Using the angle sum property in △AOB, we get
∠A + ∠O + ∠B = 180°
⇒ 7x + 1° + 60° + 70° = 180°
⇒ x = 7°
⇒ ∠A = 7 × 7° + 1° = 50°

Now, in △AOB,
∠A = 7x + 1° = 50°

Using the angle sum property in △AOB, we get
∠A + ∠O + ∠B = 180°
​⇒ 50° + 60° + ∠B = 180°
⇒ ∠B = 70°

So, in ∠AOB and ∠COD, we have
∠AOB = ∠COD                     (vertically opposite angles)
∠A = ∠D
∠B = ∠C
Thus, △AOB ~ △DOC.

Each statement alone is sufficient.

Hence, the correct answer is option (c).

Answer:

(i) Given: A model of a boat is made on the scale of 1 : 4.

$\mathrm{Scale}=\frac{\mathrm{Length} \mathrm{in} \mathrm{image}}{\mathrm{Corresponding} \mathrm{length} \mathrm{in} \mathrm{object}}$

We have, the full size of the boat has a width of 60 cm.

$$\begin{gathered} \frac{1}{4}=\frac{\mathrm{Width} \mathrm{of} \mathrm{the} \mathrm{scale} \mathrm{model}}{\mathrm{Actual} \mathrm{width} \mathrm{of} \mathrm{the} \mathrm{boat}} \\ \Rightarrow \frac{1}{4}=\frac{\mathrm{Width} \mathrm{of} \mathrm{the} \mathrm{scale} \mathrm{model}}{60} \\ \Rightarrow \mathrm{Width} \mathrm{of} \mathrm{the} \mathrm{scale} \mathrm{model}=15 \mathrm{cm} \end{gathered}$$

Hence, the correct answer is option (c).

(ii) Two similar polygons are not the mirror image of one another. They are actually of the same shape but scaled up or down.

Hence, the correct answer is option (d).

(iii) If two similar triangles have as scale factor of a : b, then their sides, medians, altitudes etc. also have the scale factor a : b. But their angles remain the same.

Answer:

(i) In the kite, both the diagonals are perpendicular to each other. So, the sticks are tied to each other at 90°.

Hence, the correct answer is option (c).

(ii) Let represent the given triangle as shown below:



In △AOB and △AOD, we have

AO = AO        (Common)

∠AOB = ∠AOD = 90°              (Diagonals are perpendicular)\

BO = OD                

∴ △AOB ~ △AOD by SAS similarity.

Hence, the correct answer is option (b).

(iii) Let two similar triangles are △ABC and △PQR with medians AD and PS respectively.

     

We have △ABC ~ △PQR

We know that the sides of similar triangles are proportional to each other and corresponding angles are equal.

$$\begin{gathered} \therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{4}{9} \\ \mathrm{And}, \angle \mathrm{B}=\angle \mathrm{Q} \end{gathered}$$

D and S are the midpoints of BC and QR respectively.

$$\begin{gathered} \therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{2\mathrm{BD}}{2\mathrm{QS}} \left[\mathrm{From} \left(1\right)\right] \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{BD}}{\mathrm{QS}} \\ \mathrm{And}, \angle \mathrm{B}=\angle \mathrm{Q} \left[\mathrm{From} \left(2\right)\right] \\ \Rightarrow △\mathrm{ABD}~△\mathrm{PQS} \left[\mathrm{By} \mathrm{SAS}\hspace{0.17em}\mathrm{similarity}\right] \\ \therefore \frac{\mathrm{BD}}{\mathrm{QS}}=\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{4}{9} \end{gathered}$$

Hence, the correct answer is option (b).

(iv) The Converse of Pythagoras theorem “In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle”.

Hence, the correct answer is option (d).

(v) The kite formed by two perpendicular sticks of length 6 cm and 8 cm is shown below:

                   

The area of the kite = Area of △ABC + Area of △ADC
                               $$\begin{gathered} =\frac{1}{2}\times \mathrm{AC}\times \mathrm{BO}+\frac{1}{2}\times \mathrm{AC}\times \mathrm{OD} \left[\because \mathrm{Area} \mathrm{of} \mathrm{triangle}=\frac{1}{2}\times b\times h\right] \\ =\frac{1}{2}\times 8\times 3+\frac{1}{2}\times 8\times 3 \\ =24 {\mathrm{cm}}^2 \end{gathered}$$

Hence, the correct answer is option (c).

Answer:

(i) Shadow A'B'C'D' of cardboard ABCD is formed on the top of the table such that quadrilateral A'B'C'D' is an enlargement of quadrilateral ABCD with scale factor 1 : 2.

Thus, quadrilateral A'B'C'D' is similar to quadrilateral ABCD and the measure of the angle does not change if the triangles are similar.

∴ ∠A' = 105°

Hence, the correct answer is option (a).

(ii) Quadrilateral A'B'C'D' is similar to quadrilateral ABCD.

∴ ∠A' = 105° and ∠C' = 70°

⇒ ∠A' + ∠C' = 105° + 70° = 175°

Hence, the correct answer is option (c).

(iii) Quadrilateral A'B'C'D' is an enlargement of quadrilateral ABCD with scale factor 1 : 2.

Thus, $\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=\frac{\mathrm{B}\text{'}\mathrm{C}\text{'}}{\mathrm{BC}}=\frac{\mathrm{C}\text{'}\mathrm{D}\text{'}}{\mathrm{CD}}=\frac{\mathrm{A}\text{'}\mathrm{D}\text{'}}{\mathrm{AD}}=\frac{2}{1}$

$$\begin{gathered} \therefore \frac{\mathrm{Perimeter} \mathrm{of} \mathrm{A}\text{'}\mathrm{B}\text{'}\mathrm{C}\text{'}\mathrm{D}\text{'}}{\mathrm{Perimeter} \mathrm{of} \mathrm{ABCD}}=\frac{2}{1} \\ \Rightarrow \mathrm{Perimeter} \mathrm{of} \mathrm{A}\text{'}\mathrm{B}\text{'}\mathrm{C}\text{'}\mathrm{D}\text{'}=2\left(\mathrm{Perimeter} \mathrm{of} \mathrm{ABCD}\right) \\ \Rightarrow \mathrm{Perimeter} \mathrm{of} \mathrm{A}\text{'}\mathrm{B}\text{'}\mathrm{C}\text{'}\mathrm{D}\text{'}=2\left(\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{AD}\right) \\ \Rightarrow \mathrm{Perimeter} \mathrm{of} \mathrm{A}\text{'}\mathrm{B}\text{'}\mathrm{C}\text{'}\mathrm{D}\text{'}=2\left(1.5+2.5+2.4+2.1\right) \mathrm{cm} \\ \Rightarrow \mathrm{Perimeter} \mathrm{of} \mathrm{A}\text{'}\mathrm{B}\text{'}\mathrm{C}\text{'}\mathrm{D}\text{'}=17 \mathrm{cm} \end{gathered}$$

Hence, the correct answer is option (d).

(iv) Quadrilateral A'B'C'D' is an enlargement of quadrilateral ABCD with scale factor 1 : 2.

Thus, the corresponding sides of the quadrilaterals is in the ratio 1 : 2.

$$\begin{gathered} \frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=\frac{\mathrm{B}\text{'}\mathrm{C}\text{'}}{\mathrm{BC}}=\frac{\mathrm{C}\text{'}\mathrm{D}\text{'}}{\mathrm{CD}}=\frac{\mathrm{A}\text{'}\mathrm{D}\text{'}}{\mathrm{AD}}=\frac{2}{1} \\ \Rightarrow \frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=\frac{2}{1} \\ \Rightarrow \mathrm{A}\text{'}\mathrm{B}\text{'}=2\times 1.5 \\ \Rightarrow \mathrm{A}\text{'}\mathrm{B}\text{'}=3 \mathrm{cm} \end{gathered}$$

Hence, the correct answer is option (b).

(v) Quadrilateral A'B'C'D' is similar to quadrilateral ABCD.

∴ ∠C' = 70° and ∠D' = 85°

⇒ ∠C' + ∠D' = 70° + 85° = 155°

Hence, the correct answer is option (c).

Answer:

(i) For figure F,
In △PQR and △STU, we have
PQ = ST
QR = TU
PR = SU

Thus, △PQR and △STU are congruent by SSS congruency.

Disclaimer:- Congruency can be proved in only figure F.

(ii) All congruent figures are similar.

Hence, the correct answer is option (b).

(iii) Converse of Thales Theorem "If a line divides any two sides of the triangle in the same ratio, then the line is parallel to the third side.

Hence, the correct answer is option (c).

(iv) Let the given situation is represented by the figure shown below:



In △ABC and △PQR, we have

∠ABC  = ∠PQR = 90°

∠ACB = ∠PRQ                              (As the angle of the sun at the same time)

⇒ △ABC ~ △PQR                        (By AA similarity)

$$\begin{gathered} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{PQ}}{\mathrm{QR}} \\ \Rightarrow \frac{5}{7}=\frac{\mathrm{X}}{14} \\ \Rightarrow \mathrm{X}=10 \mathrm{ft} \end{gathered}$$

Hence, the correct answer is option (b).

(v) The given situation can be represented in the figure shown below:


In △ABC and △ADE, we have

∠BAC = ∠DAE           (Common)

∠ACB = ∠AED = 90°

⇒ △ABC ~ △ADE                 (By AA similarity)

$$\begin{gathered} \therefore \frac{\mathrm{AC}}{\mathrm{AE}}=\frac{\mathrm{BC}}{\mathrm{DE}} \\ \Rightarrow \frac{12}{84}=\frac{2}{\mathrm{h}} \\ \Rightarrow \mathrm{h}=14 \mathrm{m} \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

Congruent figures are always similar, whereas similar figures are not necessarily congruent.

In the case of similar figures we consider only the shapes whereas, in the case of congruent triangles, we consider both the shapes and sizes of the figure.

Congruent triangles can be treated as the special case of similar triangles when the ratio of corresponding sides is equal to 1.

Thus, Statement-1 is false, Statement-2 is true.

Hence, the correct answer is option (d).

Answer:

Statement-2: RHS similarity criterion states that "If the ratio of the hypotenuse and one side of a right-angled triangle is equal to the ratio of the hypotenuse and one side of another right-angled triangle, then the two triangles are similar".

Thus, Statement-2 is true.

Statement-1: If ΔABC and ΔPQR are right triangles right angled at C and R respectively such that $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}$, then ∠B = ∠Q.

Let ΔABC and ΔPQR are right triangles right angled at C and R respectively.



∠C = ∠R = 90°

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}$

Then according to Statement-2
⇒ △ACB ~ △PRQ                 (By RHS similarity)
Thus, Statement-1 is true

So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): If in a triangle, square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side, is a right angle.

The Converse of Pythagoras theorem states that "In a triangle, the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side, is a right angle.".

Thus, Statement-2 is true.

Statement-1 (Assertion): In ΔPQR, if PQ = 12 cm, QR = 9 cm and PR = 15 cm, then ΔPQR is a right triangle right angled at Q.


Here PQ = 12 cm, QR = 9 cm and PR = 15 cm

(PR)² = 225

(PQ)² = 144

(QR)² = 81

Thus, (PR)² = (PQ)² + (QR)²

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. (Statement-2)

∴ ∠Q = 90°.
Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).





 

Answer:

Statement-2 (Reason): If the areas of two similar triangles are equal, then the triangles are congruent.

Consider $△\mathrm{ABC}~△\mathrm{PQR}$ such that $\mathrm{ar}\left(△\mathrm{ABC}\right)=\mathrm{ar}\left(△\mathrm{PQR}\right)$
Since the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$$\begin{gathered} \frac{\mathrm{Area} \mathrm{of} △\mathrm{ABC}}{\mathrm{Area} \mathrm{of} △\mathrm{PQR}}={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^2={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^2={\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)}^2 \\ \Rightarrow 1={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^2={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^2={\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)}^2 \\ \Rightarrow \mathrm{AB}= \mathrm{PQ} \\ \mathrm{BC}=\mathrm{QR} \\ \mathrm{AC}=\mathrm{PR} \end{gathered}$$
$\Rightarrow △\mathrm{ABC}\cong △\mathrm{PQR} \left[\mathrm{By} \mathrm{SSS} \mathrm{congruency}\right]$

Thus, Statement-2 is true.

Statement-1 (Assertion): In two triangles, if corresponding angles are equal then the triangles are similar.

The AA criteria of similarity states that "If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar."

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): If areas of two similar triangles are 36 cm² and 81 cm², then the ratio of their corresponding sides is 4 : 9.

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$$\begin{gathered} \frac{\mathrm{Area} \mathrm{of} \mathrm{triangle} 1}{\mathrm{Area} \mathrm{of} \mathrm{triangle} 2}={\left(\frac{\mathrm{Side} \mathrm{of} \mathrm{triangle} 1}{\mathrm{Side} \mathrm{of} \mathrm{triangle} 2}\right)}^2 \\ \Rightarrow \frac{36}{81}={\left(\frac{\mathrm{Side} \mathrm{of} \mathrm{triangle} 1}{\mathrm{Side} \mathrm{of} \mathrm{triangle} 2}\right)}^2 \\ \Rightarrow \frac{\mathrm{Side} \mathrm{of} \mathrm{triangle} 1}{\mathrm{Side} \mathrm{of} \mathrm{triangle} 2}=\frac{6}{9} \end{gathered}$$

Thus, Statement-2 is false.

Statement-1 (Assertion): ​In ΔABC, if AB = 7 cm, BC = 24 cm and AC = 25 cm, then ΔABC is a right triangle of area 84 cm².

In ΔABC, AB = 7 cm, BC = 24 cm and AC = 25 cm, then
(AC)² = 625
(BC)² = 576
(AB)² = 49

Here, (AC)² = (AB)² + (BC)²

625 = 576 + 49

If in a triangle, the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Thus, ΔABC is right angled at B.

$\begin{array}{rcl}\mathrm{Area} \mathrm{of} △\mathrm{ABC}& =& \frac{1}{2}\times \mathrm{AB}\times \mathrm{BC}\\ & =& \frac{1}{2}\times 7\times 24\\ & =& 84 {\mathrm{cm}}^2\end{array}$

Thus, ΔABC is a right triangle of area 84 cm².

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is false.

Hence, the correct answer is option (c).

Answer:

Statement-2 (Reason): The areas of two similar triangles are in the ratio of the squares of corresponding altitudes.​

Let △ABC and △PQR are two similar triangle.

$\therefore \frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{PQR}\right)}={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^2={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^2={\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)}^2$                    .....(1)

Draw AL ⏊ BC in △ABC and PM ⏊ QR in △PQR.



Now, in △ALB and △PMQ

∠ALB = ∠PMQ   (Each 90º)

∠B = ∠Q              (Given)

Thus, by AA similarity △ALB ~ △PMQ.

$\therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AL}}{\mathrm{PM}}$                          .....(2)

From (1) and (2), we have

$\therefore \frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{PQR}\right)}={\left(\frac{\mathrm{AL}}{\mathrm{PM}}\right)}^2$

So, the areas of two similar triangles are in the ratio of the squares of corresponding altitudes.​

Thus, statement-2 is true.

Statement-1 (Assertion): ΔABC and ΔDEF are similar triangles such that BC = 4 cm, EF = 5 cm and ar(ΔABC) = 64 cm², then ar(ΔDEF) = 100 cm².
​
Given: ΔABC and ΔDEF are similar triangles such that BC = 4 cm, EF = 5 cm
Since, ar(ΔABC) = 64 cm², then ar(ΔDEF) = 100 cm².
$$\begin{gathered} \therefore \frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{64}{100} \\ \Rightarrow \frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{16}{25} \end{gathered}$$

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$$\begin{gathered} \therefore \frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}={\left(\frac{\mathrm{BC}}{\mathrm{EF}}\right)}^2 \\ \Rightarrow \frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}={\left(\frac{4}{5}\right)}^2 \\ \Rightarrow \frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{16}{25} \end{gathered}$$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).

Answer:

Statement-2 (Reason): If ΔABC ~ ​ΔDEF, then $\frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{DEF}\right)}=\frac{{\mathrm{AB}}^2}{{\mathrm{DE}}^2}=\frac{{\mathrm{BC}}^2}{{\mathrm{EF}}^2}=\frac{{\mathrm{AC}}^2}{{\mathrm{DF}}^2}$.

As the areas of two similar triangles are in the ratio of the squares of corresponding sides.​

If ΔABC and ΔDEF are similar triangles, then
$\frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{DEF}\right)}=\frac{{\mathrm{AB}}^2}{{\mathrm{DE}}^2}=\frac{{\mathrm{BC}}^2}{{\mathrm{EF}}^2}=\frac{{\mathrm{AC}}^2}{{\mathrm{DF}}^2}$

Thus, statement-2 is true.
Statement-1 (Assertion): If ΔABC and ΔDEF are similar triangles such that ar(ΔABC) = 36 cm², and ar(ΔDEF) = 49 cm², then AB : DE = 6 : 7.

If ΔABC and ΔDEF are similar triangles such that ar(ΔABC) = 36 cm², and ar(ΔDEF) = 49 cm², then according to Statement-1.

$$\begin{gathered} \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{DEF}\right)}=\frac{{\mathrm{AB}}^2}{{\mathrm{DE}}^2}=\frac{{\mathrm{BC}}^2}{{\mathrm{EF}}^2}=\frac{{\mathrm{AC}}^2}{{\mathrm{DF}}^2} \\ \Rightarrow \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{DEF}\right)}=\frac{{\mathrm{AB}}^2}{{\mathrm{DE}}^2} \\ \Rightarrow \frac{36}{49}=\frac{{\mathrm{AB}}^2}{{\mathrm{DE}}^2} \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{6}{7} \end{gathered}$$

Thus, statement-1 is true.

So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

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

Thus, Statement-2 is true.

Statement-1 (Assertion): D and E are points on sides AB and AC of ΔABC such that AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x cm. If DE || BC, then x = 5.


D and E are points on sides AB and AC of ΔABC such that AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x cm and DE || BC, then according to Statement-2.

$$\begin{gathered} \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}} \\ \Rightarrow \frac{7x-4}{3x+4}=\frac{5x-2}{3x} \\ \Rightarrow 21x^2-12x=15x^2-6x+20x-8 \\ \Rightarrow 6x^2-26x+8=0 \\ \Rightarrow 6x^2-2x-24x+8=0 \\ \Rightarrow 2x\left(3x-1\right)-8\left(3x-1\right)=0 \\ \Rightarrow \left(3x-1\right)\left(2x-8\right)=0 \\ \Rightarrow x=\frac{1}{3}, 4 \end{gathered}$$

Thus, statement-1 is false.

Answer:

In the given figure let O is the intersecting point of the diagonals AC and BD.

Now, in △AOB and △COD, we have

∠AOB = ∠COD                      (Vertically opposite angle)

∠OAB = ∠OCD                      (Alternate interior angle)

∴ △AOB ~ △COD                 (By AA similarity)

$\Rightarrow \frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{BO}}{\mathrm{DO}} .....\left(1\right)$

Thus, the diagonals of a trapezium divide each other proportionally.

So, statement-2 is true.

From (1), we have

$$\begin{gathered} \Rightarrow \frac{4}{4x+2}=\frac{x+1}{2x+4} \\ \Rightarrow 4x^2+2x+4x+2=8x+16 \\ \Rightarrow 4x^2+6x-8x+2-16=0 \\ \Rightarrow 4x^2-2x-14=0 \\ \Rightarrow 2x^2-x-7=0 \end{gathered}$$

For x = 3 the equation is not satisfied so statement-1 is false.

Hence, the correct answer is option (d).