Question 4If in two triangles - ABC and - PQR-AB-QR-BC-PR-CA-PQ- thenA - PQR - CABB - PQR - ABCC - CBA - PQRD - BCA - PQR

(A) Δ PQR ∼Δ CAB Given that in Δ ABC and Δ PQR,AB/QR=BC/PR=CA/PQ.This shows that sides of one triangle are proportional to the sides of the other triangle ...

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Question 4
If in two triangles $Δ A B C a n d Δ P Q R, \frac{A B}{Q R} = \frac{B C}{P R} = \frac{C A}{P Q}$ , then

(A) $Δ P Q R \sim Δ C A B$
(B) $Δ P Q R \sim Δ A B C$
(C) $Δ C B A \sim Δ P Q R$
(D) $Δ B C A \sim Δ P Q R$

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Solution

(A) $Δ P Q R \sim Δ C A B$

Given that in $Δ A B C a n d Δ P Q R, \frac{A B}{Q R} = \frac{B C}{P R} = \frac{C A}{P Q}$ .This shows that sides of one triangle are proportional to the sides of the other triangle.
If the sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are also equal. So by SSS similarity rule, the given triangles are similar.
i.e., $Δ C A B \sim Δ P Q R$

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Similar questions

Q. Question 4
If in two and

Δ A B C a n d Δ P Q R, \frac{A B}{Q R} = \frac{B C}{P R} = \frac{C A}{P Q}

, then

(A)

Δ P Q R \sim Δ C A B

(B)

Δ P Q R \sim Δ A B C

(C)

Δ C B A \sim Δ P Q R

(D)

Δ B C A \sim Δ P Q R

Q. Question 2
If AB = QR, BC+PR and CA = PQ, then
(A)

Δ A B C ≅ Δ P Q R

(B)

Δ C B A ≅ Δ P Q R

(C)

Δ B A C ≅ Δ B C A

(D)

Δ P Q R ≅ Δ B C A

In $Δ$ ABC and $Δ$ PQR, AB = 4 cm, BC = 5 cm, AC = 6 cm and PQ = 4 cm, QR = 5 cm, PR = 6 cm, then

Q. If in two triangles

Δ A B C

and

Δ P Q R

A B = Q R, B C = P R

and

C A = P Q,

then :

Q. If in ∆ABC and ∆PQR, we have:

\frac{A B}{Q R} = \frac{B C}{P R} = \frac{C A}{P Q}

, then
(a) ∆PQR ∼ ∆CAB
(b) ∆PQR ∼ ∆ABC
(c) ∆CBA ∼ ∆PQR
(d) ∆BCA ∼ ∆PQR

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Question 4 If in two triangles ΔABC and ΔPQR,ABQR=BCPR=CAPQ, then (A) ΔPQR∼ΔCAB (B) ΔPQR∼ΔABC (C) ΔCBA∼ΔPQR (D) ΔBCA∼ΔPQR

Question 4
If in two triangles $Δ A B C a n d Δ P Q R, \frac{A B}{Q R} = \frac{B C}{P R} = \frac{C A}{P Q}$ , then

(A) $Δ P Q R \sim Δ C A B$
(B) $Δ P Q R \sim Δ A B C$
(C) $Δ C B A \sim Δ P Q R$
(D) $Δ B C A \sim Δ P Q R$