If ABC - QRP-ar ABCar QRP-94- and BC-15 cm- then find PR-

ABC≈ QRP, arc( ABC ) #160; arc( QRP ) #160; =( 3 4 #160; ) ⇒ 1 2 ( AB ) ( BC ) sin B #160; 1 2 ( QR ) ( RP ) sin R #160; = 3 4 By c ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equal Angles Subtend Equal Sides

If ABC ∼ QR...

Question

If $△ A B C \sim △ Q R P, \frac{a r (△ A B C)}{a r (△ Q R P)} = \frac{9}{4}$ , and $B C = 15 c m$ , then find $P R$ .

Open in App

Solution

Suggest Corrections

Similar questions

△ A B C \sim △ Q R P, \frac{A r (A B C)}{A r (Q R P)} = \frac{9}{4}

A B = 18 c m

B C = 15 c m

P R

Δ A B C \sim Δ Q R P, a n (\frac{Δ A B C}{Δ Q R P}) = \frac{9}{4}

A B = \frac{9}{4},

A B = 18,

B C = 15 c m

P R

Δ A B C \sim Δ Q R P, \frac{a r (Δ A B C)}{a r (Δ P Q R)} = \frac{9}{4}

\frac{20}{3} c m

△ A B C \sim △ P Q R,

\frac{a r (A B C)}{a r (P Q R)} = \frac{9}{4}

A B = 18

c m

B C = 15

c m

Q R

\frac{a r (Δ A B C)}{a r (Δ P Q R)} = \frac{9}{4}, A B

Join BYJU'S Learning Program