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.

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

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 :

(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

AB^{2} + AC^{2} = BC^{2}

⇒ (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}

∴ AB^{2} + BC^{2} = AC^{2}

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 cm^{2} and 144 cm^{2} 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.

Hence, (C) is correct option.

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 :

∆ABC ≅ ∆QRP

or ∆CBA ≅ ∆PRQ

Hence (B) is correct option.

Question 11.

ln ∆ABC, AB = AC and ∠B = 50° then ∠C equal.

(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.

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)

∠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 :

(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

AB^{2} + BC^{2} = (6√3)^{2} + (6)^{2}

= 108 + 36

= 144 = (12)^{2} = AC^{2}

Thus, AB^{2} + BC^{2} = AC^{2}

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 cm^{2}

(B) 170 cm^{2}

(C) 70 cm^{2}

(D) 100 cm^{2}

Solution :

In ∆ABC and ∆ADE

BC || DE

Thus ∠B = ∠ADE (corresponding angles)

∠C = ∠AED (corresponding angles)

⇒ 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.

**Short Answer Type Questions**

Question 1.

In Fig. if EF || BC and GE || DC then. Prove that \(\frac { AG }{ AD }\) = \(\frac { AF }{ AB }\)

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 : \(\frac { BF }{ FE }\) = \(\frac { BE }{ EC }\)

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 BC^{2} = AC × DC, then Prove that BD = BC.

Solution :

Given : In ∆ABC, AB = AC and BC^{2} = AC × DC, where D is any (RBSESolutions.com) point on side AC.

To prove : BD = BC

Proof : Given that BC^{2} = 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 : OB^{2} + OD^{2} = OA^{2} + OC^{2} (Higher Secondary Board Raj 2015)

Solution :

Given : Let ABCD is a rectangle and O is any point inside it :

AB = CD, BC = DA [opposite sides of rectangle]

To prove : OB^{2} + OD^{2} = OA^{2} + OC^{2}

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)

OA^{2} = OE^{2} + AE^{2} ….(i)

Similarly In ∆BOE, ∠BEO = 90°, (By Pythagoras Theorem)

OB^{2} = OE^{2} + EB^{2} ……(ii)

From ∆COF, ∠CFO = 90°

OC^{2} = OF^{2} + CF^{2} …..(iii)

And in ∆DOF, ∠OFD = 90°

OD^{2} = OF^{2} + DF^{2} …..(iv)

Here AE = DF (by construction) and BE = CF (by construction) ……(v)

From equation (i), (ii), (iii) and (iv)

OA^{2} + OC^{2} = OE^{2} + AE^{2} + OF^{2} + CF^{2}

= OE^{2} + DF^{2} + OF^{2} + BE^{2} [From eqn (v)] ….(vi)

OB^{2} + OD^{2} = OE^{2} + EB^{2} + OF^{2} + DF^{2} ….(vii)

∴ OA^{2} + OC^{2} = OB^{2} + OD^{2} [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.

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

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.

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]

(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]

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

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 = \(\frac { AM }{ AB }\) = \(\frac { AN }{ AD }\)

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)

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.

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.

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

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.

∵ 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.

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 2AB^{2} = 2AC^{2} + BC^{2} [CBSE 2012]

Solution :

Given : In ∆ABC

AD ⊥ BC such that BD = 3CD

To prove : 2AB^{2} = 2AC^{2} + BC^{2}

Proof : In right angled ∆ABD

AB^{2} = AD^{2} + BD^{2}

Multiplying both sides by 2

2AB^{2} = 2AD^{2} + 2BD^{2}

2AB^{2} = 2(AC^{2} – CD^{2}) + 2(3CD)^{2} [∵ AD^{2}= AC^{2} – CD^{2} ;BD = 3CD]

⇒ 2AB^{2} = 2AC^{2} – 2CD^{2} + 18CD^{2}

= 2AC^{2} + 16CD^{2}

= 2AC^{2} + (4CD)^{2}

= 2AC^{2} + (CD + 3CD)^{2}

= 2AC^{2} + (CD + BD)^{2} (∵ 3CD = BD)

= 2AC^{2} + BC^{2} (∵ BC = CD + BD)

Thus, 2AB^{2} = 2AC^{2} + BC^{2}

Question 14.

In given figure, AD is (RBSESolutions.com) median of ∆ABC and AM ⊥ BC. Prove that :

(i) AC^{2} = AD^{2} + BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)

(ii) AB^{2} = AD^{2} – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)

(iii) AC^{2} + AB^{2} = 2AD^{2} + \(\frac { 1 }{ 2 }\) BC^{2}

Solution :

Given: In ∆ABC, D is mid-point of BC since AD is median AM ⊥ BC and AC > AB.

To prove:

(i) AC^{2} = AD^{2} + BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)

(ii) AB^{2} = AD^{2} – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)

(iii) AC^{2} + AB^{2} = 2AD^{2} + \(\frac { 1 }{ 2 }\) BC^{2}

Proof : (i) In right angled ∆AMD

AD^{2} = AM^{2} + DM^{2}

AM^{2} = AD^{2} – DM^{2}

in right ∆AMC

AC^{2} = AM^{2} + MC^{2} …..(ii)

From equation (i) and (ii),

AC^{2} = (AD^{2} – DM^{2}) + MC^{2}

⇒ AC^{2} = (AD^{2} – DM^{2}) + (DM + DC)^{2} (∵ MC = DM + DC)

⇒ AC^{2} = AD^{2} – DM^{2} + DM^{2} + DC^{2} + 2DM.DC

AC^{2} = AD^{2} + DC^{2} + 2DM.DC

AC^{2} = AD^{2} + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) + 2DM·\(\frac { BC }{ 2 }\)

Thus, AC^{2} = AD^{2} + BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) …..(iii)

(ii) In right ∆AMB

AB^{2} = AM^{2} + BM^{2}

= (AD^{2} – DM^{2}) + BM^{2} [using equation (i)]

= (AD^{2} – DM^{2}) + (BD – DM)^{2}

= AD^{2} – DM^{2} + BD^{2} + DM^{2} – 2 BD.DM

AD^{2} + BD^{2} – 2BD.DM

= AD^{2} + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) – 2 × \(\frac { 1 }{ 2 }\) BC. DM

∴ AB^{2} = AD – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)

Thus, AB^{2} = AD – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) ……(iv)

Adding equation (iii) and (iv)

AB^{2} + AC^{2} = 2AD^{2} + 2 × \(\frac { 1 }{ 4 }\) BC^{2}

= 2AD^{2} + \(\frac { 1 }{ 2 }\) BC^{2}

Thus, AB^{2} + AC^{2} = 2AD^{2} + \(\frac { 1 }{ 2 }\) BC^{2}

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)

Proof : In right ∆PDQ

PQ^{2} = QD^{2} + PD^{2} (By Pythagoras Theorem)

⇒ PD^{2} = PQ^{2} – QD^{2} ….(i)

In right ∆PDR

PR^{2} = DR^{2} + PD^{2}

⇒ PD^{2} = PR^{2} – DR^{2} …..(ii)

From equation (i) and (ii)

PQ^{2} – QD^{2} = PR^{2} – DR^{2}

⇒ a^{2} – c^{2} = b^{2} – d^{2} [putting values of PQ, QD, PR and DR]

⇒ a^{2} – b^{2} = c^{2} – d^{2}

⇒ (a + b)(a – b) = (c + d)(c – d).

Question 16.

In any quadrilateral ABCD ∠A + ∠D = 90°. Prove that AC^{2} +BD^{2} = AD^{2} + BC^{2}. [NCERT Exemplar Problem]

Solution :

Given : A quad. ABCD in which, ∠A + ∠D = 90°

To Prove : AC^{2} + BD^{2} = AD^{2} + BC^{2}.

Construction : Produce AB and DC intersect (RBSESolutions.com) each other at E.

Proof : ∠A + ∠D = 90°

In triangle ADE

∠A + ∠D + ∠E = 180°

⇒ 90° + ∠E = 180°

⇒ ∠E = 180° – 90° = 90°

In right triangle BEC

BC^{2} = BE^{2} + CE^{2}

(By Pythagoras Theorem)

AD^{2} = AF^{2} + DE^{2} …..(ii)

Adding equation (i) and (ii)

BC^{2} + AD^{2} = BE^{2} + AE^{2} + CE^{2} + DE^{2} …..(iii)

In right ∆AEC,

AC^{2} = AE^{2} + CE^{2} …(iv)

In right angled triangle BED

BD^{2} = BE^{2} + DE^{2} ……(v)

From equation (iv) and (v)

AC^{2} + BD^{2} = BE^{2} + AE^{2} + CE^{2} + DE^{2} …..(vi)

From equations (iii) and (iv)

BC^{2} + AD^{2} = AC^{2} + BD^{2}

⇒ AC^{2} + BD^{2} = AD^{2} + BC^{2}.

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]

Solution :

Given :

XY || AC

In ∆BXY and ∆BAC

∠BXY = ∠BAC (corresponding angle)

∠BYX = ∠BCA (corresponding angle)

We hope the given RBSE Solutions for Class 10 Maths Chapter 11 Similarity Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 11 Similarity Additional Questions, drop a comment below and we will get back to you at the earliest.

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