If in two triangles - ABC and - PQR-AB-QR-BC-PR-CA-PQ- then -1 mark-a - PQR - CABb - PQR - ABCc - CBA - PQRd - BCA - PQR

The correct option is (a). nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; ...

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Standard X

Mathematics

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If in two tri...

Question

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 [1 mark]

(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

The correct option is (a). [1 mark]

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

Choose the correct answer in each of the following:

$If AB = QR, BC = RP and CA = PQ then which of the following holds? (a) Δ A B C ≅ Δ P Q R (b) Δ C B A ≅ Δ P Q R (c) Δ C A B ≅ Δ P Q R (d) Δ B C A ≅ Δ P Q R$

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

Q. If AB = QR, BC = RP and CA = PQ, then which of the following holds?
(a) ∆ABC ≅ ∆PQR
(b) ∆CBA ≅ ∆PQR
(c) ∆CAB ≅ ∆PQR
(d) ∆BCA ≅ ∆PQR

Q. 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

Q. 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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If in two triangles ΔABC and ΔPQR,ABQR=BCPR=CAPQ, then [1 mark] (a) ΔPQR∼ΔCAB (b) ΔPQR∼ΔABC (c) ΔCBA∼ΔPQR (d) ΔBCA∼ΔPQR

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 [1 mark]

(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) $Δ B C A \sim Δ P Q R$