# Maharashtra Board Class 10 Math Part 2 Solution Chapter 1 Practice Set 1.3 Similarity

## Maharashtra Board Class 10 Math Part – 2 Solution Chapter 1 Practice Set 1.3 – Similarity

Balbharati Maharashtra Board Class 10 Math Part – 2 Solution Chapter 1: Similarity. Marathi or English Medium Students of Class 10 get here Similarity full Exercise Solution.

 Std Maharashtra Class 10 Subject Math Part 2 Solution Chapter Similarity Practice Set 1.3

#### Practice set 1.3

(1) In figure 1.55, ABC = 75°,

EDC = 75° state which two triangles are similar and by which test? Also write the similarity of these two triangles by a proper one to one correspondence.

Solution: In the adjacent fig, it is given that ∠ABC = ∠CDE = 75°

∴In ∆ABC and ∆CDE

∠ABC ≅∠CDE

∠DCE is common.

∴ ∆ABC ~ ∆CDE by AA test

(2) Are the triangles in figure 1.56 similar? If yes, by which test?

Solution: In the adjacent fig, PQ = 6, QR = 8, PR = 10, LM = 3, MN = 4, LN = 5

In ∆PQR & ∆LMN

∴ PQ/LM = 6/3 = 2/1

QR/MN = 8/4 = 2/1, PR/LN = 10/5 = 2/1

∴ PQ/LM = QR/MN = PR/LN

∴∆PQR ~ ∆LMN by SSS test

(3) As shown in figure 1.57, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of smaller pole due to sunlight is 6 m then how long will be the shadow of the bigger pole at the same time?

Solution: In the adjacent triangles PR⊥ to BQ  AC⊥BC

PR = 4, QR = 6, AC = 8, BC = x

Now,

In ∆PQR & ∆ABC

∠PRQ = ∠ACB = 90°

∠QPR = ∠BAC [∵ Angles made by sunlight on top are congruent]

∴∆PQR ~ ∆ABC by AA test

By property of similar triangles we can say that.

∴ PR/AC = QR/BC or, 4/8 = 6/x or, x = 12

(4) In ∆ABC, AP⊥BC, BQ⊥AC

B – P – C, A – Q – C then prove that,

∆CPA ~∆CQB

If AP = 7, BQ = 8, BC = 12 then find AC.

Solution: In the adjacent triangles if it is given that,

AP⊥BC, BQ⊥AC

B – P – C, A – Q – C

In ∆CPA & ∆CQB

∠APC = ∠CQB = 90°

∠BCQ = ∠ACP [Common angle of two triangle]

∴∆CPA ~ ∆CQB by AA test

Now, AP = 7, AB = 8, BC = 12 given

∴By property of similar triangles are can say that AP/BQ = AC/BC

Or, 7/8 = AC/12

Or, AC = 21/2 = 10.5

(5) Given: In trapezium PQRS, side PQ||side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ

Solution: PQ||SR, AR = 5AP —- (i)

Now, in ∆ASR & ∆APQ

∠SAR = ∠PAQ [Opposite vertically angle]

∠ASR = ∠AQP [Alternate angle ∵ PQ||SR]

∴∆ASR ~ ∆APQ by AA test

∴By property of similar triangle.

AR/AP = AS/AQ = SR/PQ also, AR/AP = SR/PQ

Or, 5AP/AP = AS/AQ – From equation (i) or, 5AP/AP = SR/PQ – from equation (ii)

Or, AS = 5AQ Prove/ Or, SR = 5PQ proved

(6) In trapezium ABCD, (Figure 1.60) side AB||Side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then find OD.

Solution: AB||DC, AB = 20, DC = 6, OB = 15

∆AOB & ∆DOC

∠AOB = ∠DOC [Vertically opposite angles]

∠ABO = ∠CDO [Alternate angles ∵AB||DC]

∴ ∆ADB ~ ∆DOC by AA test

∴ By property of similar triangles we can say that,

AB/DC = OB/OD = OA/OC

∴ AB/DC = OB/OD

Or, 20/6 = 15/OD

Or, OD = 15×6/20 = 9/2 = 4.5

(7) ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE×BE = CE×TE

Solution: In the adjacent ABCD is a parallelogram.

∵ ABCD is a parallelogram.

∴ AD || BC & AB||DC

BT is an extention of seg AB as mentioned in question.

∴In ∆DEC & ∆BET

∠BET = ∠DEC [Vertically opposite angles]

∠TBE = ∠DCE [Alternate angles ∵ AT||DC]

∴∆DEC ~ ∆BET by AA test

∴by property of similar triangles we can say that,

DE/TE = CE/BE = DC/TB

Or, DE/TE = CE/BE

Or, DE × BE = CE×TE Proved

(8) In the figure, seg AC and seg BD intersect each other in point P and AP/CP = BP/DP. Prove that,

AP/CP = BP/DP. Prove that,

∆ABP ~∆CDP

Solution: AP/CP = BP/DP [Given]

∴In∆APB & ∆CPD

AP/CP = BP/DC [Given]

∠APB = ∠CPD [Vertically opposite angles]

∴∆APB ~ ∆CPD by SAS test

(9) In the figure, in ∆ABC, point D on side BC such that,

Prove that, CA2 = CB×CD

Solution: In the adjacent fig it is given that,

∠ACB = ∠ACD [Common angle of two triangle]

∴∆ABC ~ ∆ADC by AA test.

∴by the property similar we can say that