Question

The area of the trapezium $PQRS$ is approximately:

• A. $8.4c{m}^2$
• B. $5.75c{m}^2$
• C. $5.25c{m}^2$
• D. $4.2c{m}^2$

Answer 1

Answer: (D)

$4.2c{m}^2$

According to the property of trapezium, PQ must be parallel to RS.

So, AS = RB

$PS\sin 60=QR\sin 30PS=2.3\times \frac{\sin 30}{\sin 60}PS=1.3$

From the figure, it is clear that,

RS = PQ-(PA+QB)

$RS=PQ-(PS\cos 60+QR\cos 30)RS=5-(1.3\cos 60+2.3\cos 30)RS=2.3$

In the trapezium PQRS,

$SA=PS\sin 60SA=1.3\sin 60SA=1.1$

Now, area of trapezium PQRS

$=\frac{1}{2}\times \left(Sumoflengthofparallelsides\right)\times height=\frac{1}{2}\times (2.3+5)\times 1.1=4.2$

So, correct answer is option D.