RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions

RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 11 Similarity Additional Questions.

Rajasthan Board RBSE Class 10 Maths Chapter 11 Similarity Additional Questions

Multiple Choice Questions
Question 1.
If ∆ABC ~ ∆DEF and AB = 10 cm, DE = 8 cm, then ar. (∆ABC) : ar. (∆DEF) will (RBSESolutions.com) be :
(A) 25 : 16
(B) 16 : 25
(C) 4 : 5
(D) 5 : 4
Solution :
∵ ∆ABC ~ ∆DEF
⇒ \(\frac { ar.\triangle ABC }{ ar.\triangle DEF } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } }\)
⇒ \(\frac { ar.\triangle ABC }{ ar.\triangle DEF } =\frac { { 10 }^{ 2 } }{ { 8 }^{ 2 } }\)
⇒ \(\frac { ar.\triangle ABC }{ ar.\triangle DEF } =\frac { 100 }{ 64 }\)
⇒ \(\frac { ar.\triangle ABC }{ ar.\triangle DEF } =\frac { 25 }{ 16 }\)
= 25 : 16
Thus, choice (A) is correct.

RBSE Solutions

Question 2.
In triangles ABC and DEF, AB = FD and ∠A = ∠D. Two triangles (RBSESolutions.com) are congruent by SAS criterion if
(A) BC = EF
(B) AC = DE
(C) AC = EF
(D) BC = DE
Solution :
For congruence by SAS Rule AC = DE
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 1
Thus (B) is correct option.

Question 3.
In figure, if DE || AB then AD = 2 cm, DC = 3 cm and BE = 3 cm and (RBSESolutions.com) then value of CE will be :
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 2
(A) 4.5 cm
(B) 2 cm
(C) 18 cm
(D) 5 cm
Solution :
Since DE || AB, according to question here AD = 2 cm, DC = 3 cm and BE = 3 cm
Thus, By Basic Prop. Theorem
\(\frac { CD }{ AD }\) = \(\frac { CE }{ BE }\)
⇒ CE = \(\frac { BE\times DC }{ AD } \)
∴ CE = \(\frac { 3\times 3 }{ 2 } \)
Thus (A) is correct.

Question 4.
If in ∆ABC, ∠A = 90°, BC = 25 cm, AC = 7 cm, then (RBSESolutions.com) AB will be:
(A) 34 cm
(B) 24 cm
(C) 16 cm
(D) 28 cm
Solution :
In right angled triangle ∆CAB, by Pythagoras theorem
AB2 + AC2 = BC2
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 3
⇒ (AB)2 + (7)2 = (25)2
⇒ (AB)2 = (25)2 – (7)2
⇒ (AB)2 = 625 – 49
⇒ (AB)2 = 576
AB = \(\sqrt { 576 }\) = 24 cm
Thus, (B) is correct.

Question 5.
In ∆ABC if AB = 5 cm, BC = 12 cm (RBSESolutions.com) and AC = 13 cm, then ∠B will be:
(A) 120°
(B) 60°
(C) 90°
(D) 45°
Solution :
In ∆ABC,
(AB)2 + (BC)2 = (5)2 + (12)2
= 25 + 144 = 169
= (AC)2
∴ AB2 + BC2 = AC2
By converse of Pythagoras Theorem
∆ABC is right angle triangle, In which
∠B = 90°
Hence, choice (C) is correct.

Question 6.
∆ABC and ∆PQR are two similar triangles (RBSESolutions.com) whose areas are 100 cm2 and 144 cm2 and height of ∆ABC is 6 cm. then height of ∆PQR will be :
(A) 12 cm
(B) 6.3 cm
(C) 7.2 cm
(D) 4.8 cm
Solution :
We know that ratio of areas of two similar triangles is equal to ratio of square of corresponding heights. Let corresponding height of ∆PQR is x m.
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 4
Hence, (C) is correct option.

RBSE Solutions

Question 7.
In ∆ABC and ∆DEF if \(\frac { AB }{ DE }\) = \(\frac { BC }{ FD }\) then they will be (RBSESolutions.com) similar if : (NCERT Exemplar Problem)
(A) ∠B = ∠E
(B) ∠A = ∠D
(C) ∠B = ∠D
(D) ∠A = ∠F
Solution :
Option (C) is correct.

Question 8.
If in ∆DEF and ∆PQR if ∠D = ∠Q and ∠R = ∠E then which one from the following is correct? (NCERT Exemplar Problem)
(A) \(\frac { EF }{ PR }\) = \(\frac { DF }{ PQ }\)
(A) \(\frac { DE }{ PQ }\) = \(\frac { EF }{ RP }\)
(A) \(\frac { DE }{ QR }\) = \(\frac { DF }{ PQ }\)
(A) \(\frac { EF }{ RP }\) = \(\frac { DE }{ QR }\)
Solution :
(B) is correct.

Question 9.
From the following which is (RBSESolutions.com) not the criterion of congruence.
(A) SAS
(B) ASA
(C) SSA
(D) SSS
Solution :
(C) is correct option.

Question 10.
If AB = QR, BC = PR and CA = PQ then :
(A) ∆ABC ≅ ∆PQR
(B) ∆CAB ≅ ∆PQR
(C) ∆BAC ≅ ∆RPQ
(D) ∆PQR ≅ ∆BCA
Solution: According to question :
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 5
∆ABC ≅ ∆QRP
or ∆CBA ≅ ∆PRQ
Hence (B) is correct option.

Question 11.
ln ∆ABC, AB = AC and ∠B = 50° then ∠C equal.
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 6
(A) 40°
(B) 50°
(C) 80°
(D) 130°
Solution :
∵ Opposite angles of (RBSESolutions.com) equal sides are equal.
∴ ∠B = ∠C = 50 cm
Hence, option (B) is correct.

RBSE Solutions

Question 12.
In ∆ABC, BC = AB and ∠B = 80° then ∠A equals.
(A) 80°
(B) 40°
(C) 50°
(D) 100°
Solution :
∴ Opposite angles of equal sides are equal. ∠A = ∠C = x (Let)
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 7
∠x + ∠x + ∠B = 180°
∠x + ∠x + 80° = 180°
2∠x = 180° – 80° = 100° ⇒ x = ∠A = 50°
Hence option (C) is correct

Question 13.
In ∆PQR, ∠R = ∠P and QR = 4 cm and PR = 5 cm then (RBSESolutions.com) length of PQ is :
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 8
(A) 4 cm
(B) 5 cm
(C) 2 cm
(D) 2.5 cm
Solution:
∴ ∠P = ∠R
∴ QR = PQ
⇒ PQ = QR = 4 cm
Thus option (A) is correct.

Question 14.
Choose correct answer (RBSESolutions.com) and give reason. In ∆ABC, if AB = 6\(\sqrt { 3 }\) cm, AC = 12 cm and BC = 6 cm, then ∠B is :
(A) 120°
(B) 60°
(C) 90°
(D) 45°
Solution :
AC = 12 cm
AB = 6\(\sqrt { 3 }\) cm
BC = 6 cm
AB2 + BC2 = (6√3)2 + (6)2
= 108 + 36
= 144 = (12)2 = AC2
Thus, AB2 + BC2 = AC2
By converse Pythagoras theorem
In ∆ABC ∠B = 90°
Thus (C) is correct option.

Question 15.
In ∆ABC if BC || DE and DE = 4 cm and BC = 8 cm and ar. (∆ADE) = 25 sq cm then ar. (∆ABC) will be :
(A) 110 cm2
(B) 170 cm2
(C) 70 cm2
(D) 100 cm2
Solution :
In ∆ABC and ∆ADE
BC || DE
Thus ∠B = ∠ADE (corresponding angles)
∠C = ∠AED (corresponding angles)
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 9
⇒ By AA similarity criterion
∆ABC ~ ∆ADE
Thus, areas of two similar triangles are equal (RBSESolutions.com) to ratio of squares of their corresponding sides.
∴ \(\frac { ar.\triangle ABC }{ ar.\triangle ADE } =\frac { { BC }^{ 2 } }{ { DE }^{ 2 } }\)
⇒ \(\frac { ar.\triangle ABC }{ 25 } =\frac { { 8 }^{ 2 } }{ { 4 }^{ 2 } }\)
ar.(∆ ABC) = \(\frac { 64\times 25 }{ 16 } \)
100 sq. cm.
Hence, option (D) is correct.

RBSE Solutions

Short Answer Type Questions
Question 1.
In Fig. if EF || BC and GE || DC then. Prove that \(\frac { AG }{ AD }\) = \(\frac { AF }{ AB }\)
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 10
Solution :
In ∆ADC
GE || DC
∴ \(\frac { AG }{ AD }\) = \(\frac { AE }{ AC }\) (By Basic Prop. Theorem)
Thus, in ∆ABC
∵ FE || BC
∴ \(\frac { AE }{ AC }\) = \(\frac { AF }{ AB }\)
From equation (i) and (ii)
\(\frac { AG }{ AD }\) = \(\frac { AF }{ AB }\)

Question 2.
In figure DE || AC and DF || AE. (RBSESolutions.com) Prove that : \(\frac { BF }{ FE }\) = \(\frac { BE }{ EC }\)
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 11
Solution :
Given : D is any point on side AB of ∆ABC and E and F are two points on side BC. line segment DF, DE and AE are drawn.
To prove that : \(\frac { BF }{ FE }\) = \(\frac { BE }{ EC }\)
Proof : In ∆BCA,
DE || AC (given)
\(\frac { BE }{ EC }\) = \(\frac { BD }{ DA }\) …..(i) (By Basic Prop. Theorem)
Again In ∆BEA, DF || AE (given)
\(\frac { BF }{ FE }\) = \(\frac { BD }{ DA }\) …..(ii) (By Basic Prop. Theorem)
From equation (i) and (ii)
\(\frac { BF }{ FE }\) = \(\frac { BE }{ EC }\)

Question 3.
In ∆ABC if AB = AC and a point D lies of AC such that BC2 = AC × DC, then Prove that BD = BC.
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 12
Solution :
Given : In ∆ABC, AB = AC and BC2 = AC × DC, where D is any (RBSESolutions.com) point on side AC.
To prove : BD = BC
Proof : Given that BC2 = AC × DC
⇒ \(\frac { BC }{ DC }\) = \(\frac { AC }{ BC }\) …..(i)
Now in ∆ABC and ∆BDC
∠C = ∠C (common angle)
and \(\frac { BC }{ DC }\) = \(\frac { AC }{ BC }\) [from (i)]
Thus, by SAS similarity criterion
∆ABC ~ ∆BDC
⇒ \(\frac { AC }{ BC }\) = \(\frac { AB }{ BD }\) …..(ii)
Given, AB = AC
Thus, \(\frac { AC }{ BC }\) = \(\frac { AC }{ BD }\)
⇒ BC = BD.

Question 4.
O is any point ¡n the interior of (RBSESolutions.com) rectangle ABCD prove that : OB2 + OD2 = OA2 + OC2 (Higher Secondary Board Raj 2015)
Solution :
Given : Let ABCD is a rectangle and O is any point inside it :
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 13
AB = CD, BC = DA [opposite sides of rectangle]
To prove : OB2 + OD2 = OA2 + OC2
Construction : Draw a line passing through O and (RBSESolutions.com) parallel to BC which cuts AB and CD at E and F respectively.
[∵ BC || EF, ∴ ∠AEF = ∠ABC = 90° and ∠AEO = ∠AEF.]
In ∆AOE, ∠AEO = 90°, (By Pythagoras Theorem)
OA2 = OE2 + AE2 ….(i)
Similarly In ∆BOE, ∠BEO = 90°, (By Pythagoras Theorem)
OB2 = OE2 + EB2 ……(ii)
From ∆COF, ∠CFO = 90°
OC2 = OF2 + CF2 …..(iii)
And in ∆DOF, ∠OFD = 90°
OD2 = OF2 + DF2 …..(iv)
Here AE = DF (by construction) and BE = CF (by construction) ……(v)
From equation (i), (ii), (iii) and (iv)
OA2 + OC2 = OE2 + AE2 + OF2 + CF2
= OE2 + DF2 + OF2 + BE2 [From eqn (v)] ….(vi)
OB2 + OD2 = OE2 + EB2 + OF2 + DF2 ….(vii)
∴ OA2 + OC2 = OB2 + OD2 [From equation (vi) and (vii)]

Question 5.
The shadow of a vertical pillar of length 6 m. (RBSESolutions.com) On earth is 4 m where as at the same time shadow of a tower is 28 m. Find the height of the tower.
Solution :
Given : The shadow DE = 4 of 6 m. long pillar CD is obtained. At the same time shadow BE 28 m of a tower of height (say h) is obtained. To find : height (h) of tower AB.
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 14
Calculation : ∆ABE and ∆CDE are similar
∴ \(\frac { AB }{ CD }\) = \(\frac { BE }{ DE }\)
⇒ \(\frac { h }{ 6 }\) = \(\frac { 28 }{ 4 }\)
⇒ h = \(\frac { 28 }{ 4 }\) × 6
h = 42 m
Hence, height of tower = 42 m

RBSE Solutions

Question 6.
AD and PM are respectively (RBSESolutions.com) medians of ∆ABC and ∆PQR respectively whereas ∆ABC ~ ∆PQR
Prove that : \(\frac { AB }{ PQ }\) = \(\frac { AD }{ PM }\)
Solution :
Given : ∆ABC and ∆PQR are similar triangles In which AD and PM are medians of ∆ABC and ∆PQR respectively.
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 15
To prove : \(\frac { AB }{ PQ }\) = \(\frac { AD }{ PM }\)
Proof : ∆ABC and ∆PQR are similar
\(\frac { AB }{ PQ }\) = \(\frac { BC }{ QR }\) ….(i)
∠Q = ∠B (∆ABC ~ ∆PQR)
∵ AD, is median of ∆ABC
∴ BD = \(\frac { 1 }{ 2 }\) BC ⇒ BC = 2 BD
and PM, is median of ∆PQR.
∴ QM = \(\frac { 1 }{ 2 }\) QR ⇒ QR = 2 QM
From equation (i),
\(\frac { AB }{ PQ }\) = \(\frac { 2BD }{ 2QM }\)
⇒ \(\frac { AB }{ PQ }\) = \(\frac { BD }{ QM }\) …..(ii)
Now, comparing ∆ABD and ∆PQM
\(\frac { AB }{ PQ }\) = \(\frac { BD }{ QM }\)
∠B = ∠Q
∴ By S-A-S similarity criterion
∆ABD ~ ∆PQM
\(\frac { AB }{ PQ }\) = \(\frac { AD }{ PM }\) (Corresponding sides (RBSESolutions.com) of similar triangles are proportional)

Question 7.
In figure, CD and RS are medians of ∆ABC and ∆PQR respectively. If ∆ABC ~ ∆PQR then Prove that : [Higher Secondary Board Raj. 2014]
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 16
(i) ∆ABC ~ ∆PQR
(ii) \(\frac { CD }{ RS }\) = \(\frac { AB }{ PQ }\)
Solution :
(i) Given : Two similar triangles ABC and PQR CD ⊥ AB and RS ⊥ PQ.
To prove : ∆ADC ~ ∆PSR
Proof : \(\frac { CA }{ RP }\) = \(\frac { AB }{ PQ }\)
⇒ \(\frac { CA }{ RP }\) = \(\frac { 2AD }{ 2PS }\) (Since D and S. are (RBSESolutions.com) medians of AB and PQ)
Now, in ∆ACD and ∆PRS,
\(\frac { CA }{ RP }\) = \(\frac { AD }{ PS }\)
and ∠A = ∠P (∵ ∆ABC ~ ∆PQR)
∠ADC = ∠PSR = 90°
Thus, by AA similarity Criterion.
∆ADC ~ ∆PSR.

(ii) Given :
∆ADC ~ ∆PSR.
To prove : \(\frac { CD }{ RS }\) = \(\frac { AB }{ PQ }\)
Proof : Given that ∆ABC and ∆PQR are similar. So
∠A = ∠P ….(i) (∵ Corresponding angle S of (RBSESolutions.com) similar triangles are same.)
\(\frac { CA }{ RP }\) = \(\frac { AB }{ PQ }\) = \(\frac { CB }{ RQ }\) ….(ii)
No, In ∆CAD and ∆RPS,
∠A = ∠P (given)
∠CDA = ∠RSP = 90°
By A-A similarity criterion
∆CDA ~ ∆RSP
\(\frac { CA }{ RP }\) = \(\frac { CD }{ RS }\)
From equation (ii) and (iii)
\(\frac { CD }{ RS }\) = \(\frac { AB }{ PQ }\)

Question 8.
In given fig. DE || BC and AD : DB = 7 : 5 then find \(\frac { ar.\triangle ABC }{ ar.\triangle DEF }\) [CBSE – 2012]
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 17
Solution :
Given : DE || BC and AD : DB = 7 : 5
In ∆ADE and ∆ABC,
∠ADE = ∠ABC (Corresponding angle)
∠AED = ∠ACB (Corresponding angle)
∴ ∆ADE ~ ∆ABC (AA Similarity criterion)
⇒ \(\frac { AD }{ AB }\) = \(\frac { DE }{ BC }\) = \(\frac { AE }{ AC }\) ….(i)
[Corresponding sides of (RBSESolutions.com) similar triangles re proportional]
But AD : DB = 7 : 5
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 18
Putting value of \(\frac { AD }{ AB }\) in equation (i)
\(\frac { AD }{ AB }\) = \(\frac { DE }{ BC }\) = \(\frac { AE }{ AC }\) = \(\frac { 12 }{ 7 }\)
⇒ \(\frac { DE }{ BC }\) = \(\frac { 12 }{ 7 }\)
In ∆DFE and ∆CFB,
∠DFE = ∠CBF (Alternate angle)
∠DFE = ∠CFB (Vertically opposite angle)
∴ ∆DFE ~ ∆CFB (AA similarity criterion)
⇒ \(\frac { ar.\triangle DFE }{ ar.\triangle CFB } =\frac { { DE }^{ 2 } }{ { CB }^{ 2 } }\)
⇒ \(\frac { ar.\triangle DFE }{ ar.\triangle CFB }\) = \({ \left( \frac { DE }{ BC } \right) }^{ 2 }\)
⇒ \(\frac { ar.\triangle DFE }{ ar.\triangle CFB }\) = \({ \left( \frac { 12 }{ 7 } \right) }^{ 2 }\) [Using equation (ii)]
⇒ \(\frac { ar.\triangle DFE }{ ar.\triangle CFB }\) = \(\frac { 144 }{ 49 }\)
Thus, \(\frac { ar.\triangle DFE }{ ar.\triangle CFB }\) = \(\frac { 144 }{ 49 }\)

Question 9.
In figure LM || CB and LN || CD then. Prove that = \(\frac { AM }{ AB }\) = \(\frac { AN }{ AD }\)
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 19
Solution :
In ∆ABC,
ML || BC (given)
\(\frac { AM }{ AB }\) = \(\frac { AL }{ LC }\) ……(i) (By Basic Prop. Theorem)
Again in ∆ADC,
LN || DC (given)
\(\frac { AN }{ ND }\) = \(\frac { AL }{ LC }\) ……(ii) (By Basic prop. Theorem)
From equation (i) and (ii)
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 20

Question 10.
Points E and F lie on sides PQ and PR of any ∆PQR. From the (RBSESolutions.com) following, for each case, find is EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, and PF = 0.36 cm
Solution :
In ∆PQR, two point E and F Lie on sides PQ and PR respectively.
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 21
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 22
EF, is not parallel to QR.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm.
\(\frac { PE }{ QE }\) = \(\frac { 4 }{ 4.5 }\) = \(\frac { 40 }{ 45 }\) = \(\frac { 8 }{ 9 }\) …..(i)
and \(\frac { PF }{ RF }\) = \(\frac { 8 }{ 9 }\)
From equation (i) and (ii)
\(\frac { PE }{ QE }\) = \(\frac { PF }{ RF }\) (By converse of Basic Prop. Theorem)
∴ EF || QR
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm.
EQ = PQ – PE
= 1.28 – 0.18 = 1.10 cm
FR = PR – PF
2.56 – 0.36 = 2.20 cm
\(\frac { PE }{ EQ }\) = \(\frac { 0.18 }{ 1.10 }\) = \(\frac { 18 }{ 110 }\) = \(\frac { 9 }{ 55 }\) …..(i)
and \(\frac { PF }{ FR }\) = \(\frac { 0.36 }{ 2.20 }\) = \(\frac { 36 }{ 220 }\) = \(\frac { 9 }{ 55 }\) …..(ii)
From equation (i) and (ii)
\(\frac { PE }{ EQ }\) = \(\frac { PF }{ FR }\) (By converse of Basic Prop. Theorem)
EF || QR

RBSE Solutions

Question 11.
Prove that ratio of areas of two (RBSESolutions.com) similar triangles is equal to ratio of their corresponding medians.
Solution :
Given : Two similar triangles ABC and DEF. AX and DY are respectively medians of sides BC and EF.
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 23
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 24
∵ Ratio of areas of two similar (RBSESolutions.com) triangles is equal to ratio of squares of their corresponding sides.
\(\frac { ar.\triangle ABC }{ ar.\triangle DEF }\) = \(\frac { { AB }^{ 2 } }{ { DE }^{ 2 } }\) = \(\frac { { AX }^{ 2 } }{ { DY }^{ 2 } }\)

Question 12.
Sides AB and AC and median AD of a triangle are respectively proportional to sides PQ and PR and median PM of another triangle. Show that ∆ABC ~ ∆PQR.
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 25
Solution :
Given : In two triangles ABC and ∆PQR D is mid-point of BC and M is mid-point of QR.
and \(\frac { AB }{ PQ }\) = \(\frac { AC }{ PR }\) = \(\frac { AD }{ PM }\) ……(i)
To prove : ∆ABC ~ ∆PQR
Construction : Produce AD upto E such that AD = DE. Join BE and CE and produce PM upto N such that PM = MN. Join QN and NR.
Proof : Diagonals AE and BC of (RBSESolutions.com) quadrilateral ABEC bisect each other at point D.
∴ quadrilateral ABEC is a parallelogram.
∴ BE = AC …(ii)
Similarly PQNR is a parallelogram
∴ QN = PR …..(iii)
Dividing equation (ii) by (iii)
\(\frac { BE }{ QN }\) = \(\frac { AC }{ PR }\) …..(iv)
Now, \(\frac { AD }{ PM }\) = \(\frac { 2AD }{ 2PM }\) = \(\frac { AD }{ PM }\) = \(\frac { AE }{ PN }\) …(v)
In equation (i), (iv) and (v),
\(\frac { AB }{ PQ }\) = \(\frac { BE }{ QN }\) = \(\frac { AE }{ PN }\)
Thus, in ∆ABE and ∆PQN
\(\frac { AB }{ PQ }\) = \(\frac { BE }{ QN }\) = \(\frac { AE }{ PN }\)
∴ ∆ABE ~ ∆PQN (By SSS)
∴ ∠BAE = ∠QPN …(vi)
Similarly
∆AEC ~ ∆PNR
∠EAC = ∠NPR …(vii)
Adding equation (vi) and (vii)
∠BAE + ∠EAC = ∠QPN + ∠NPR
⇒ ∠BAC = ∠QPR
Now, in ∆ABC and ∠PQR
\(\frac { AB }{ PQ }\) = \(\frac { AC }{ PR }\) [From equation (i)]
∠A = ∠P
∴ By SAS similarity criterion
∆ABC ~ ∆PQR

Question 13.
From vertex A of ∆ABC, a perpendicular is (RBSESolutions.com) drawn on side BC at point D such that DB = 3 CD. Show that 2AB2 = 2AC2 + BC2 [CBSE 2012]
Solution :
Given : In ∆ABC
AD ⊥ BC such that BD = 3CD
To prove : 2AB2 = 2AC2 + BC2
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 26
Proof : In right angled ∆ABD
AB2 = AD2 + BD2
Multiplying both sides by 2
2AB2 = 2AD2 + 2BD2
2AB2 = 2(AC2 – CD2) + 2(3CD)2 [∵ AD2= AC2 – CD2 ;BD = 3CD]
⇒ 2AB2 = 2AC2 – 2CD2 + 18CD2
= 2AC2 + 16CD2
= 2AC2 + (4CD)2
= 2AC2 + (CD + 3CD)2
= 2AC2 + (CD + BD)2 (∵ 3CD = BD)
= 2AC2 + BC2 (∵ BC = CD + BD)
Thus, 2AB2 = 2AC2 + BC2

Question 14.
In given figure, AD is (RBSESolutions.com) median of ∆ABC and AM ⊥ BC. Prove that :
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 27
(i) AC2 = AD2 + BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)
(ii) AB2 = AD2 – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)
(iii) AC2 + AB2 = 2AD2 + \(\frac { 1 }{ 2 }\) BC2
Solution :
Given: In ∆ABC, D is mid-point of BC since AD is median AM ⊥ BC and AC > AB.
To prove:
(i) AC2 = AD2 + BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)
(ii) AB2 = AD2 – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)
(iii) AC2 + AB2 = 2AD2 + \(\frac { 1 }{ 2 }\) BC2
Proof : (i) In right angled ∆AMD
AD2 = AM2 + DM2
AM2 = AD2 – DM2
in right ∆AMC
AC2 = AM2 + MC2 …..(ii)
From equation (i) and (ii),
AC2 = (AD2 – DM2) + MC2
⇒ AC2 = (AD2 – DM2) + (DM + DC)2 (∵ MC = DM + DC)
⇒ AC2 = AD2 – DM2 + DM2 + DC2 + 2DM.DC
AC2 = AD2 + DC2 + 2DM.DC
AC2 = AD2 + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) + 2DM·\(\frac { BC }{ 2 }\)
Thus, AC2 = AD2 + BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) …..(iii)

(ii) In right ∆AMB
AB2 = AM2 + BM2
= (AD2 – DM2) + BM2 [using equation (i)]
= (AD2 – DM2) + (BD – DM)2
= AD2 – DM2 + BD2 + DM2 – 2 BD.DM
AD2 + BD2 – 2BD.DM
= AD2 + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) – 2 × \(\frac { 1 }{ 2 }\) BC. DM
∴ AB2 = AD – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)
Thus, AB2 = AD – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) ……(iv)
Adding equation (iii) and (iv)
AB2 + AC2 = 2AD2 + 2 × \(\frac { 1 }{ 4 }\) BC2
= 2AD2 + \(\frac { 1 }{ 2 }\) BC2
Thus, AB2 + AC2 = 2AD2 + \(\frac { 1 }{ 2 }\) BC2

Question 15.
In ∆PQR, PD ⊥ QR such that (RBSESolutions.com) lie on line segment QR. If PQ = a, PR = b, QD = c and DR = d Prove that :
(a + b)(a – b) = (c + d)(c – d). [NCERT Exemplar Problem]
Solution :
Given : PD ⊥ QR. PQ = a, PR = b, QD and DR = d.
To prove : (a + b)(a – b) = (c + d)(c – d)
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 28
Proof : In right ∆PDQ
PQ2 = QD2 + PD2 (By Pythagoras Theorem)
⇒ PD2 = PQ2 – QD2 ….(i)
In right ∆PDR
PR2 = DR2 + PD2
⇒ PD2 = PR2 – DR2 …..(ii)
From equation (i) and (ii)
PQ2 – QD2 = PR2 – DR2
⇒ a2 – c2 = b2 – d2 [putting values of PQ, QD, PR and DR]
⇒ a2 – b2 = c2 – d2
⇒ (a + b)(a – b) = (c + d)(c – d).

RBSE Solutions

Question 16.
In any quadrilateral ABCD ∠A + ∠D = 90°. Prove that AC2 +BD2 = AD2 + BC2. [NCERT Exemplar Problem]
Solution :
Given : A quad. ABCD in which, ∠A + ∠D = 90°
To Prove : AC2 + BD2 = AD2 + BC2.
Construction : Produce AB and DC intersect (RBSESolutions.com) each other at E.
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 29
Proof : ∠A + ∠D = 90°
In triangle ADE
∠A + ∠D + ∠E = 180°
⇒ 90° + ∠E = 180°
⇒ ∠E = 180° – 90° = 90°
In right triangle BEC
BC2 = BE2 + CE2
(By Pythagoras Theorem)
AD2 = AF2 + DE2 …..(ii)
Adding equation (i) and (ii)
BC2 + AD2 = BE2 + AE2 + CE2 + DE2 …..(iii)
In right ∆AEC,
AC2 = AE2 + CE2 …(iv)
In right angled triangle BED
BD2 = BE2 + DE2 ……(v)
From equation (iv) and (v)
AC2 + BD2 = BE2 + AE2 + CE2 + DE2 …..(vi)
From equations (iii) and (iv)
BC2 + AD2 = AC2 + BD2
⇒ AC2 + BD2 = AD2 + BC2.

Question 17.
In given figure, line segment XY is parallel (RBSESolutions.com) to side AC of ∆ABC and divides the triangle in two equal parts. Find ratio \(\frac { AX }{ AB }\) [CBSE – 2012]
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 30
Solution :
Given :
XY || AC
In ∆BXY and ∆BAC
∠BXY = ∠BAC (corresponding angle)
∠BYX = ∠BCA (corresponding angle)
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 31
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions 32

RBSE Solutions

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