Construct $\Delta PQR$ in which $QR = 4\,cm$,$\angle Q = 45^{\circ}$ and $PQ + PR = 8\,cm$. Write the steps of construction.
(N/A) Steps of construction:
$1$. Draw a line segment $QR = 4\,cm$.
$2$. At point $Q$,construct an angle $\angle XQR = 45^{\circ}$.
$3$. From the ray $QX$,cut off a line segment $QD = 8\,cm$ (which is equal to $PQ + PR$).
$4$. Join $DR$.
$5$. Construct the perpendicular bisector of $DR$,which intersects $QD$ at point $P$.
$6$. Join $PR$. Thus,$\Delta PQR$ is the required triangle.
$1$. Draw a line segment $QR = 4\,cm$.
$2$. At point $Q$,construct an angle $\angle XQR = 45^{\circ}$.
$3$. From the ray $QX$,cut off a line segment $QD = 8\,cm$ (which is equal to $PQ + PR$).
$4$. Join $DR$.
$5$. Construct the perpendicular bisector of $DR$,which intersects $QD$ at point $P$.
$6$. Join $PR$. Thus,$\Delta PQR$ is the required triangle.
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