# Free Triangles 03 Practice Test - 10th Grade

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 _____.

A.

4.41 cm

B.

8.4 cm

C.

4.2 cm

D.

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=14

Let the height of bigger triangle be  x

(2.1)2x2=14x=4×(2.1)2
=(2×2.1)=4.2 cm

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

A.

True

B.

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.

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? A.

True

B.

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

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. A.

True

B.

False

#### SOLUTION

Solution : B

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

STQR ...... (converse of BPT)

Therefore, PST=RQR and PTQ=PTR

Therefore PSTPQR ...... (AA similarity)

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

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

#### 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

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

A.

2 : 3

B.

4 : 9

C.

81 : 16

D.

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

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. A.

True

B.

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.

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

MC = 5 cm

B.

AM = 5 cm

C.

MB = 5 cm

D.

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

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 ΔABCΔEBD. A. True
B. False

#### SOLUTION

Solution : A In ΔABCandΔEBD,
BEBA=BDBC=26=13 and DBE=ABC.
By SAS similarity,
ΔABC and ΔEBD are similar. In the given figure, ABCPQR. Then, area of  ABCarea of  PQR equals

A.

AB2PQ2

B.

BC2QR2

C.

AC2PR2

D. All of the above.

#### SOLUTION

Solution : D

We are given two triangles ABC and PQR such that ABCPQR.

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 ABCPQR)
and AMB=PNQ=90.

So, ABMPQN
( By AA similarity criterion)
AMPN=ABPQ  (2)
Also, ABCPQR   (Given)
So, ABPQ=ACPR=BCQR  (3)

From (1) and (3), we get,
area of (ABC)area of (PQR)=AB×AMPQ×PN                    =AB×ABPQ×PQ                    =AB2PQ2

Now using (3), we get,
area of ABCarea of PQR=AB2PQ2=BC2QR2=AC2PR2