Practice Set 1.3

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

###### Practice Set 1.3

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