Question

Find the area of given figure, if AB = EF = 12 cm, CD = 10 cm and PQ = QR = 5 cm.

• A. 55 $c{m}^2$
• B. 60 $c{m}^2$
• C. 110 $c{m}^2$
• D. 220 $c{m}^2$

Answer 1

The correct option is C 110 $c{m}^2$

PQ is perpendicular to AB and CD.
$\Rightarrow$ AB || CD
$\Rightarrow$ ABCD is a trapezium

Similarly,
QR is perpendicular to CD and EF.
$\Rightarrow$ CD || EF
$\Rightarrow$ CDEF is a trapezium

Therefore,
Area of given figure
= Area of trapezium ABDC + Area of trapezium CDFE

We know that
Area of trapezium
$=\frac{1}{2}\times \left(Sumofparallelsides\right)\times height$

Area of trapezium ABDC
$=\frac{1}{2}\times (AB+CD)\times PQ$
$=\frac{1}{2}\times (12+10)\times 5=\frac{1}{2}\times 22\times 5$
= 11$\times$5
= 55 $c{m}^2$

Area of trapezium CDFE
$=\frac{1}{2}\times (EF+CD)\times QR$
$=\frac{1}{2}\times (12+10)\times 5=\frac{1}{2}\times 22\times 5$
= 11$\times$5
= 55 $c{m}^2$

Therefore, Area of given figure = 55 + 55 = 110 $c{m}^2$