Question

Construct a $\Delta PQR$ in such a way that $\angle PQR={60}^{\circ }$, $\angle PRQ={30}^{\circ }$ and QR = 8 cm. Construct a rectangle, the area of which is equal to the area of $\Delta PQR$. What are the dimensions of the rectangle.

• A. (3.6, 4)
• B. (9, 8)
• C. (4.5, 7)
• D. (9.1, 5)

Answer 1

The correct option is A

(3.6, 4)

Method of Construction:

(1) Draw a line segment QT and then the part QR = 8 cm is cut from QT.
(2) Draw $\angle RQX$ equal to ${30}^{\circ }$ at Q of QR and $\angle QRY$ equal to ${60}^{\circ }$ at R of QR. Let QX and RY intersect at P. Then $\Delta PQR$ is the required triangle.
(3) Draw AB through P parallel to QR
(4) Draw perpendicular bisector DE of QR, which intersects AB at E.
(5) Cut EF from EB equal to DR.
(6) Join R and F

$\therefore$ DEFR is the required rectangle. On measuring we find that the dimensions are (3.6, 4) approximately.