Question

if triangle ABC similar triangle QRP,

$\frac{ar(abc}{ar (qrp)} = \frac{9}{4}$
And BC = 15cm then find PR​

Answer 1

$\huge\bf\blue{Hey\:Mate!!!}$

We know that ratio of area of similar triangle is equal to the square of their corresponding sides.

$\frac{ar(ABC)}{ar(QRP)} = {( \frac{BC}{PR} )}^{2} \\ \\ \frac{9}{4} = {( \frac{15}{PR}) }^{2} \\ \\taking \: square \: roots \: on \: both \: sides\\ \\ \frac{3}{2} = \frac{15}{PR} \\ \\3PR = 30\\ \\ PR = 10$__________________________

@ItsMysteriousGirl✒️

Answer 2

αɳรωεɾ 

Side PR will be 10 cm .

รσℓµƭเσɳ

As given that ∆ABC ≈ ∆ QRP

$\frac{ar(abc)}{ar(qrp)} = \frac{9}{4}$

and Side BC = 15cm .

As we know that if triangle are similar then the ratio of similar triangle will be equal to the square of the corresponding sides.

So applying this we got :-

$\frac{ar(abc)}{ar(qrp)} = \frac{9}{4} = \frac{( {bc)}^{2} }{( {pr)}^{2} }$

So we got:-

${( \frac{bc}{pr} )}^{2} = \frac{9}{4}$

$\frac{bc}{pr} = \sqrt{ \frac{9}{4} }$

As we know that √9 = 3 and √4 = 2 . That is equal to -

$\frac{bc}{pr} = \frac{3}{2}$

Now putting the value of BC we got :-

$\frac{15}{pr} = \frac{3}{2}$

3 PR = 15 × 2

PR. =. 5 × 2

PR. = 10 cm .

hope it helps