# Free Triangles 03 Practice Test - 10th Grade

### Question 1

The areas of two similar triangles are 12 cm2 and 48 cm2. If the height of the smaller triangle is 2.1 cm, then the corresponding height of the bigger triangle is _____.

4.41 cm

8.4 cm

4.2 cm

0.525 cm

#### SOLUTION

Solution :C

The ratio of the area of two similar triangles is equal to the ratio of the squares of their corresponding heights

So,

(Height of smaller △)2(Height of bigger △)2=1248=14Let the height of bigger triangle be x

∴(2.1)2x2=14⇔x=√4×(2.1)2

=(2×2.1)=4.2 cm

### Question 2

All congruent figures are similar but the similar figures need not be congruent.

True

False

#### SOLUTION

Solution :A

All congruent figures are similar but the similar figures need not be congruent as in case of similar figures only shape is considered whereas, in the case of congruent figures, both shape and sizes are considered. Hence, the statement is correct.

### Question 3

In both the figures given below, all of the respective sides are equal and all interior angles are 90∘. Are the figures shown below similar?

True

False

#### SOLUTION

Solution :A

Both the figures are similar as all the corresponding angles are equal and all the corresponding sides are in the same ratio as all the sides of both the figures are equal.

Ratio of corresponding sides = 4.22.1=2

### Question 4

In the figure below, if the line segment ST is parallel to line segment QR such that PSSQ=PTTR. The provided data is not sufficient to prove that triangles PQR and PST are similar.

True

False

#### SOLUTION

Solution :B

PSSQ=PTTR ...... (given)

ST∥QR ...... (converse of BPT)

Therefore, ∠PST=∠RQR and ∠PTQ=∠PTR

Therefore △PST∼△PQR ...... (AA similarity)

Therefore, the information provided in the question is sufficient to prove that the triangles are similar.

### Question 5

In given figure △ ABC and △ DEF are similar, BC=3cm, EF=4cm, and area of triangle ABC=54cm2 find the area of △ DEF

#### SOLUTION

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

So 916 =54area of △DEF

area of △ DEF = 96cm2

### Question 6

If the sides of two similar triangles are in the ratio of 4 : 9, then the areas of these triangles are in the ratio ____.

2 : 3

4 : 9

81 : 16

16 : 81

#### SOLUTION

Solution :D

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

Given, the sides of two triangles are in the ratio of 4 : 9.

Therefore, the ratio of areas of these triangles is equal to (49)2, which is, 1681.

### Question 7

A farmer had three land plots as shown. He has to divide the plots equally among 2 of his sons. He gives Area a and Area b to one of his son and third Area c to his second son. Did he give equal land plots to both of his son?

Enter True if he has given equal area else enter false.

True

False

#### SOLUTION

Solution :A

Pythagoras theorem states that a2+b2=c2

So he did correct i.e. he divided the land equally among his two sons.

### Question 8

In the given figure, ∠ABC = 90∘ and BM is a median, AB = 8 cm and BC = 6 cm. Then, choose the correct option(s).

MC = 5 cm

AM = 5 cm

MB = 5 cm

AC = 5 cm

#### SOLUTION

Solution :A, B, and C

In triangle ABC, by applying pythagoras rule,

AC2=AB2+BC2

AC2=82+62=64+36

AC=√64+36=√100=10 cm

Since, M is the mid-point of AC,

AM=MC=AC2=102=5 cm

We know that in a right angled triangle, mid-point of the hypotenuse is always at equidistance from all the three vertices of the triangle.

⇒ AM = MC = MB = 5 cm

### Question 9

In ΔABC, AB=BC=6cm. If a circle is drawn with center at B and radius 2 cm which intersects AB and BC at E and D, then ΔABC∼ΔEBD.

#### SOLUTION

Solution :A

In ΔABCandΔEBD,

BEBA=BDBC=26=13 and ∠DBE=∠ABC.

By SAS similarity,

ΔABC and ΔEBD are similar.

### Question 10

In the given figure, △ABC∼△PQR. Then, area of △ ABCarea of △ PQR equals

AB2PQ2

BC2QR2

AC2PR2

#### SOLUTION

Solution :D

We are given two triangles ABC and PQR such that △ABC∼△PQR.

For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles ABC and PQR respectively, as shown below.

Now,

area of △ABC=12×BC×AM and area of △PQR=12×QR×PN

So,

area of △ABCarea of △PQR=12×BC×AM12×QR×PN=BC×AMQR×PN⋯(1)

Now, in △ABM and △PQN,

∠B=∠Q (As △ABC∼△PQR)

and ∠AMB=∠PNQ=90∘.

So, △ABM∼△PQN

( By AA similarity criterion)

∴AMPN=ABPQ ⋯(2)

Also, △ABC∼△PQR (Given)

So, ABPQ=ACPR=BCQR ⋯(3)

From (1) and (3), we get,

area of (ABC)area of (PQR)=AB×AMPQ×PN =AB×ABPQ×PQ =AB2PQ2Now using (3), we get,

area of △ABCarea of △PQR=AB2PQ2=BC2QR2=AC2PR2