In the figure, $ O Q: P Q=3: 4 $ and perimeter of $ \Delta P O Q=60 \mathrm{~cm} $. Determine $ P Q, Q R $ and $ O P $."

Given:

In the figure, \( O Q: P Q=3: 4 \) and perimeter of \( \Delta P O Q=60 \mathrm{~cm} \).

To do:
We have to determine \( P Q, Q R \) and \( O P \).

 Solution:

In the figure,

$OQ : PQ = 3:4$

Perimeter of $\triangle POQ = 60\ cm$

Let $OQ = 3x$ and $PQ = 4x$

In right angled triangle $OPQ$,

$OP^2 = OQ^2 + PQ^2$

$= (3x)^2 + (4x)^2$

$= 9x^2 + 16x^2$

$= 25x^2$

$= (5x)^2$

$\Rightarrow OP = 5x$

$OQ + QP + OP = 60\ cm$

$3x + 4x + 5x = 60$

$12x = 60$

$x = 5$

$PQ = 4x = 4 \times 5 = 20\ cm$

$QR = 2 OQ = 2(3x) = 6 \times 5 = 30\ cm$

$OP = 5x = 5 \times 5 = 25\ cm$

Therefore, $PQ=20\ cm, QR= 30\ cm$ and $OP=25\ cm$.

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Updated on: 10-Oct-2022

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