Question

PQRS is a trapezium such that PQ || RS and PQ = 10 cm, QR = 5 cm, RS = 8 cm and PS = 5 cm, then area of PQRS is

• A.
  $12\sqrt{3}c{m}^2$
• B.
  $18\sqrt{6}c{m}^2$
• C.
  $24\sqrt{3}c{m}^2$
• D.
  $12\sqrt{6}c{m}^2$

Answer 1

The correct option is B

$18\sqrt{6}c{m}^2$
PQRS is a trapezium in which PQ = 10 cm, QR = 5 cm, RS = 8 cm and PS = 5 cm.

By construction, $RT||PSandRU\perp PQ.$
Therefore, PTRS is a parallelogram.

Also, $\Delta RQT$ is an isosceles triangle.



$\therefore Areaof\Delta RQT=\frac{QT}{4}\sqrt{4(QR{)}^2-Q{T}^2}\Rightarrow \frac{1}{2}\times QT\times RU=\frac{QT}{4}\sqrt{4(QR{)}^2-Q{T}^2}\Rightarrow RU=\frac{1}{2}\sqrt{4(5{)}^2-(2{)}^2}\Rightarrow RU=\frac{1}{2}\sqrt{100-4}=\frac{\sqrt{96}}{2}=\frac{4\sqrt{6}}{2}=2\sqrt{6}cm\therefore \text{Area of trapezium PQRS}=\frac{1}{2}\times (PQ+RS)\times RU=\frac{1}{2}\times (10+8)\times 2\sqrt{6}=18\sqrt{6}c{m}^2$ Hence, the correct answer is option (b).