Practice Set 1.3 Similarity Class 10th Mathematics Part 2 MHB Solution

Practice Set 1.3

1. In figure 1.55, ∠ABC = 75°, ∠EDC = 75° state which two triangles are similar and by…
2. Are the triangles in figure 1.56 similar? If yes, by which test?
3. As shown in figure 1.57, two poles of height 8 m and 4 m are perpendicular to the…
4. In ΔABC, AP ⊥ BC, BQ ⊥ AC B- P-C, A-Q - C then prove that, ΔCPA ~ ΔCQB. If AP = 7, BQ =…
5. Given : In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR =…
6. In trapezium ABCD, (Figure 1.60) side AB || side DC, diagonals AC and BD intersect in…
7. □ ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T.…
8. In the figure, seg AC and seg BD intersect each other in point P and ap/cp = bp/dp…
9. In the figure, in ΔABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA^2…

Practice Set 1.3

Question 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.

Answer:

With one- to-one correspondence ABC ↔ EDC

∵ ∠ABC ≅ ∠EDC = 75°

∠ACB ≅ ∠ECD (Is common in both the triangles ABC and EDC)

⇒ Δ ABC~Δ EDC ………(By AA Test)

Question 2.

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

Answer:

In Δ PQR and Δ LMN

And 

And 

⇒ 

⇒ ΔPQR~ΔLMN …………(By SSS Similarity Test)

Question 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?

Answer:

∵ the shadows are measured at the same time

⇒ angle of of elevation will be equal for both the pole

⇒ Δ PQR~Δ ABC ………(By AA Test)

⇒ 

⇒ BC

⇒ x = 

⇒ x = 12 m

Question 4.

In ΔABC, AP ⊥ BC, BQ ⊥ AC B- P-C, A-Q - C then prove that, ΔCPA ~ ΔCQB. If AP = 7, BQ = 8, BC = 12 then find AC.

Answer:

From fig.

⇒ ∠ APC≅ ∠ BQC (∵ AP⊥ BC and BQ⊥ AC)

⇒ Also, ∠ ACP≅ ∠ BCQ (Common)

⇒ Δ CPA~Δ CQB (By AA Test)

⇒ 

⇒ AC = 

⇒AC = 

⇒AC = 10.5

Question 5.

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

Answer:

Given that, AR = 5AP and AS = 5AQ

⇒  = 5 ………(1)

And  = 5……….(2)

⇒ 

And, ∠ SAR≅ ∠ QAP …… (opposite angles)

⇒ Δ SAR ~ Δ QAP …………(SAS Test of similarity)

⇒  (corresponding sides are proportional)

But,  = 5

⇒  = 5

⇒ SR = 5PQ

Question 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.

Answer:

In Δ AOB and ΔCOD

⇒ ∠ AOB≅ ∠ COD (opposite angles)

⇒ ∠ CDO≅ ∠ ABO (Alternate angles ∵ AB||DC)

⇒ Δ AOB ~ Δ COD (By AA Test)

⇒  (corresponding sides are proportional)

⇒ OD = 

⇒ OD = 

⇒ OD = 4.5

Question 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.

Answer:

In Δ CED and ΔBET

⇒ ∠ CED≅ ∠ BET (opposite angles)

⇒ ∠ CDE≅ ∠ BTE (Alternate angles)

(∵ AB||DC ⇒ BT||DC, as BT is extension to AB)

⇒ Δ CED ~ Δ BET (By AA Test)

⇒  (corresponding sides are proportional)

⇒ DE× BE = CE×TE

Question 8.

In the figure, seg AC and seg BD intersect each other in point P and  Prove that, ΔABP ~ ΔCDP

Answer:

In Δ APB & Δ CPD

⇒  ……(Given)

And, ∠APB = ∠DPC (vertically opposite angles)

⇒ Δ APB ~ Δ CPD (By SAS Test)

Question 9.

In the figure, in ΔABC, point D on side BC is such that, ∠BAC = ∠ADC.

Prove that, CA2 = CB × CD

Answer:

In Δ BAC & Δ ADC

⇒ ∠ BAC ≅ ∠ ADC ……(Given)

And, ∠ACB ≅ ∠DCA ……(common)

⇒ Δ BAC ~ Δ ADC (By AA Test)

⇒  (corresponding sides are proportional)

⇒ CA2 = CB×CD