Question

If triangle ABC ~ triangle ar(ABC)=16cm^2 and ar(PQR)=81cm^2,AB=2cm.Find PQ

Answer 1

16÷81=AB^2÷PQ^2
16÷81=2^2÷PQ^2
PQ^2=4×81÷16
PQ^2=81÷4
PQ=
$\sqrt{81 \div 4}$
PQ=9/2

Answer 2

Answer:

PQ = 4.5 cm

Step-by-step explanation:

Given ∆ABC ~∆PQR,

ar(∆ABC) = 16 cm² and

ar(∆PQR)=81 cm²,

AB = 2 cm ,

PQ = ?

/*We know the theorem:

The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

$\frac{(PQ)^{2}}{(AB)^{2}}=\frac{A_{2}^{2}}{A_{1}^{2}}$

$\implies \big(\frac{PQ}{AB}\big)^{2} = \frac{81 \:cm^{2}}{16\:cm^{2}}$

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

$\implies \frac{PQ}{AB}=\frac{9}{4}$

$\implies \frac{PQ}{2}=\frac{9}{4}$

$\implies PQ = \frac{9 \times 2}{4}$

$\implies PQ = \frac{9}{2}\\=4.5 \: cm$

Therefore,

PQ = 4.5 cm

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