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
