In figure OQ PQ = 3 4 and perimeter of Δ POQ = 60 cm

SOLUTION

Let OP = $x$, then using the ratio relationship we get,

$PQ=\frac{3}{4}x$

Using pythogreas theorem to obtain OQ:

$\Rightarrow x^2+\frac{3}{4}{x}^2=O{Q}^2$

$\Rightarrow x^2+\frac{9}{16}{x}^2=O{Q}^2$

$\Rightarrow x^2\frac{25}{16}=O{Q}^2$

Getting their square root:

$\therefore OQ=\sqrt{(}\frac{25}{16}{x}^2)=\frac{5}{4}x$

Using the given Perimeter to get the unknown :

$\frac{5}{4}x+\frac{3}{4}x+x=60cm$

$\frac{12}{4}x=60cm$

$3x=60cm$

$x=20cm$

$\frac{3}{4}$ × 20 = 15cm

OP = 20cm

PQ = 15cm