Triangles Test

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Triangles Test - 12

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Question 1
1 / -0

In ΔABC and ΔPQR, ∠B
=∠Q, ∠R=∠C and AB
=2QR.
Then,the triangles are

A
similar but not congruent.

B
neither congruent nor similar.

C
congruent as well as similar.

D
congruent but not similar.

Solution

In ΔABC and ΔPQR,
∠B =∠Q, ∠R =∠C and AB = 2PQ
Then the triangles are similar, by AA similarity rule, but not congruent because for congruency, sides should also be equal.
Question 2
1 / -0

In the given figure if BP||CF, DP||EF, then AD:DE is equal to

A
3:4

B
1:4

C
2:3

D
1:3

Solution

since BP \(\| CF\)
Then, \(\frac{ AP }{ PF }=\frac{ AB }{ BC }[ Using\) Thales Theorem]
\(\Rightarrow \frac{ AP }{ PF }=\frac{2}{6}=\frac{1}{3}\)
Again, since DP || EF,
Then, \(\frac{ AP }{ PF }=\frac{ AD }{ DE }[ Using\) Thales Theorem]
\(\Rightarrow \frac{ AD }{ DE }=\frac{1}{3}\)
\(\Rightarrow AD : DE =1: 3\)
Question 3
1 / -0

If ΔABC ∼ ΔDEF then which of the following is true?

A
BC.EF = AC.FD

B
BC.DE = AB.FD

C
BC.DE = AB.EF

D
AB.EF = AC.DE

Solution

If \(\Delta A B C \sim \Delta D E F\)
then \(B C\). \(D E=A B\). \(E F \Rightarrow \frac{B C}{E F}=\frac{A B}{D E}\) (corresponding sides are in proportion)
Here according to given condition, BC. DE =AB .EF
Question 4
1 / -0

In the given figure if DE||BC, then x is equal to

A
15

B
19

C
11

D
17

Solution

Given: \(D E \| B C\)
\(\therefore \frac{ AD }{ DB }=\frac{ AE }{ EC } \Rightarrow \frac{4}{x-4}=\frac{8}{3 x-19}\) by using Thale's theorem
\(\Rightarrow 12 x-76=8 x-32\)
\(\Rightarrow 4 x=44\)
\(\Rightarrow x=11\)
Question 5
1 / -0

∠ADE = ∠ABC, then CE is equal to

A
5

B
3

C
2

D
4.5

Solution

In \(\Delta ABC\) and \(\Delta ADE\)
\(\angle ADE =\angle ABC\) [ Given ]
\(\angle A =\angle A\) [ Common ]
\(\therefore \Delta ABC \sim \Delta ADE\) [ AA Similarity]
\(\therefore \frac{ AD }{ DB }=\frac{ AE }{ EC }\)
\(\Rightarrow \frac{2}{3}=\frac{3}{ EC }\)
\(\Rightarrow EC =4.5 cm\)
Question 6
1 / -0

In the given figure AD:DB = 1:3, AE:EC = 1:3 and BF:FC = 1:4, then

A
AD||FC

B
DE||BC

C
AE||DF

D
AD||FE

Solution

Given: \(\frac{ AD }{ DB }=\frac{1}{3}\) and \(\frac{ AE }{ EC }=\frac{1}{3}\)
Therefore, in \(\Delta ABC , \frac{ AD }{ DB }=\frac{ AE }{ EC }\)
\(\therefore\) DE \(\|\) BC [Using Thales Theorem]
Here we are not considering \(BF : FC =1: 4\)
Question 7
1 / -0

In the given figure XY||BC. If AX = 3cm, XB = 1.5cm and BC = 6cm, then

A
4 cm

B
3 cm

C
6 cm

D
4.5 cm

Solution

Since \(X Y \| B C,\) then using Thales theorem,
\(\Rightarrow \frac{ AX }{ AB }=\frac{ XY }{ BC }\)
\(\Rightarrow \frac{3}{4.5}=\frac{X Y}{6}\)
\(\Rightarrow X Y=4 cm\)
Question 8
1 / -0

In the adjoining figures RS||DB||PQ. If CP = PD = 11 and DR = RA = 3. Then,

A
x = 16, y = 8

B
x = 12, y = 10

C
x = 10, y = 7

D
x = 14, y = 6

Solution

In \(\Delta CDB , PQ \| DB [ Given ]\)
\(\therefore \frac{ CP }{ CD }=\frac{ PQ }{ BD }\) [Using Thales Theorem]
\(\Rightarrow \frac{11}{22}=\frac{8}{x} \Rightarrow x=\frac{8 \times 22}{11}=16\)
Again, In \(\Delta ABD ,\) RS \(\| DB\) [Given]
\(\therefore \frac{ AR }{ AD }=\frac{ RS }{ BD }[ Using\) Thales Theorem]
\(\Rightarrow \frac{3}{6}=\frac{y}{16}\)
\(\Rightarrow y=\frac{3 \times 16}{6}=8\)
\(\therefore x=16, y=8\)
Question 9
1 / -0

It is given that ΔABC ∼ ΔDFE, ∠A = 30^o,∠C = 40^o, AB = 5cm, AC = 8cm, and DF = 7.5cm. Then, the following is true.

A
∠F = 40^o, DE = 12cm

B
∠F = 110^o, DE = 12cm

C
∠D = 30^o, EF = 12cm

D
∠D = 110^o, EF = 12cm

Solution

In \(\Delta ABC , \angle A +\angle B +\angle C =180^{\circ}\)
\(\Rightarrow 30^{\circ}+40^{\circ}+\angle B =180^{\circ}\)
\(\Rightarrow \angle B =110^{\circ}\)
Since \(\Delta A B C \sim \Delta D F E\)
therefore, \(\angle B =\angle F =110^{\circ}\)
Also \(\frac{ DF }{ DE }=\frac{ AB }{ AC }\)
\(\Rightarrow \frac{7.5}{ DE }=\frac{5}{8}\)
\(\Rightarrow DE =12 cm\)
Question 10
1 / -0

In the given figure, DE||BC. AB = 15cm, BD = 6cm, AC = 25cm, then AE is equal to

A
18 cm

B
20 cm

C
15 cm

D
10 cm

Solution

since \(D E \| B C,\) then using Thales theorem, \(\Rightarrow \frac{ AB }{ DB }=\frac{ AC }{ EC }\)
\(\Rightarrow \frac{15}{6}=\frac{25}{ EC }\)
\(\Rightarrow E C=10 cm\)
Now, \(A E=A C-E C=25-10=15 cm\)
Question 11
1 / -0

In the adjoining figure P and Q are points on the sides AB and AC respectively of ΔABC such that AP = 3.5 cm, PB = 7cm, AQ = 3 cm, QC = 6 cm and PQ = 4.5 cm. The measure of BC is equal to

A
9 cm

B
12.5 cm

C
15 cm

D
13.5 cm

Solution

In \(\Delta ABC\)
\(\Rightarrow \frac{ AQ }{ QC }=\frac{ AP }{ PB } \Rightarrow \frac{3}{6}=\frac{3.5}{7} \Rightarrow \frac{1}{2}=\frac{1}{2}\)
since \(\frac{ AQ }{ QC }=\frac{ AP }{ PB }\)
therefore, \(QP \| BC\)
\(\therefore \frac{ AQ }{ AC }=\frac{ QP }{ BC }\)
\(\Rightarrow \frac{3}{9}=\frac{4.5}{ BC }\)
\(\Rightarrow BC =13.5 cm\)
Question 12
1 / -0

In the given figures the measures of ∠P and ∠R are respectively

A
50^o, 40^o

B
40^o, 50^o

C
30^o, 20^o

D
20^o, 30^o

Solution

In \(\Delta ABC , \angle A +\angle B +\angle C =180^{\circ}\)
\(\Rightarrow \angle A +30^{\circ}+20^{\circ}=180^{\circ}\)
\(\Rightarrow A =130^{\circ}\)
Again, in \(\Delta ABC\) and \(\Delta QRP\),
\(\frac{ AB }{ QR }=\frac{ CA }{ PQ }\)
\(\Rightarrow \frac{45}{5}=\frac{63}{7}\)
\(\Rightarrow \frac{9}{1}=\frac{9}{1}\)
since, Sides of \(\Delta ABC\) and \(\Delta QRP\) are proportional, and \(\angle A =\angle Q\)
therefore, by SAS Similarity rule, \(\Delta ABC \sim \Delta QRP\)
\(\therefore \angle A =\angle Q , \angle B =\angle R\) and \(\angle C =\angle P\)
\(\Rightarrow \angle P =20^{\circ}\)
and \(\angle R =30^{\circ}\)