Construct a triangle $ P Q R $ with side $ Q R=7 \mathrm{~cm}, P Q=6 \mathrm{~cm} $ and $ \angle P Q R=60^{\circ} $. Then construct another triangle whose sides are $ 3 / 5 $ of the corresponding sides of $ \triangle P Q R $.
Given:
A triangle \( P Q R \) with side \( Q R=7 \mathrm{~cm}, P Q=6 \mathrm{~cm} \) and \( \angle P Q R=60^{\circ} \).
To do:
We have to construct a triangle \( P Q R \) with side \( Q R=7 \mathrm{~cm}, P Q=6 \mathrm{~cm} \) and \( \angle P Q R=60^{\circ} \). Then construct another triangle whose sides are \( 3 / 5 \) of the corresponding sides of \( \triangle P Q R \).
Solution:
Steps of construction:
(i) Draw a line segment $QR = 7\ cm$.
(ii) At $Q$ draw a ray $QX$ making an angle of $60^o$ and cut off $PQ = 6\ cm$. Join $PR$.
(iii) Draw a ray $QY$ making an acute angle and cut off five equal parts.
(iv) Join $Q_5R$ and through $Q_3$, draw $Q_3S$ parallel to $Q_5R$ which meets $QR$ at $S$.
(v) Through $S$, draw $ST\ \parallel\ RP$ meeting $PQ$ at $T$.
$QST$ is the required triangle.
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