Question
Construct a right$-$angled triangle in which: $QP = QR$ and hypotenuse $PR = 7 \ cm$
✓
Answer
In $\triangle PQR,$
$QP = QR ....($given$)$
$\Rightarrow \angle QPR = \angle QRP$
Since hypotenuse $PR = 7\ cm, \angle PQR = 90^\circ $
$\therefore \angle QPR + \angle QRP = 90^\circ $
$\Rightarrow \angle QPR = \angle QRP = 45^\circ $
Steps:
$1$. Draw $PR = 7\ cm.$
$2$. Draw a ray $PT$ such as $\angle RPT = 45^\circ $ and ray $RS$ such as $\angle PRS = 45^\circ $
$3$. Ray $RS$ and ray $PT$ meets at $Q.$
Thus, $\text{PQR}$ is the required triangle.
$QP = QR ....($given$)$
$\Rightarrow \angle QPR = \angle QRP$
Since hypotenuse $PR = 7\ cm, \angle PQR = 90^\circ $
$\therefore \angle QPR + \angle QRP = 90^\circ $
$\Rightarrow \angle QPR = \angle QRP = 45^\circ $
Steps:
$1$. Draw $PR = 7\ cm.$
$2$. Draw a ray $PT$ such as $\angle RPT = 45^\circ $ and ray $RS$ such as $\angle PRS = 45^\circ $
$3$. Ray $RS$ and ray $PT$ meets at $Q.$
Thus, $\text{PQR}$ is the required triangle.
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