Question

Find the area of figure, if HD = 4 cm, GE = 5 cm, AF = BF = HF = 6 cm and CH = 7 cm.

• A. 97 $c{m}^2$
• B. 103 $c{m}^2$
• C. 111 $c{m}^2$
• D. 119 $c{m}^2$

Answer 1

The correct option is C 111 $c{m}^2$

Area of given figure
= Area of $△$BFA
+ Area of trapezium BFHC
+ Area of $△$CHD + Area of $△$AED

Area of triangle = $\frac{1}{2}$$\times$base$\times$height

Area of trapezium BFHC
= $\frac{1}{2}$$\times$(sum of parallel sides)$\times$height
= $\frac{1}{2}$$\times$(BF+CH)$\times$FH
= $\frac{1}{2}$$\times$(6+7)$\times$6
= $\frac{1}{2}$$\times$13$\times$6 = 13$\times$3 = 39 $c{m}^2$

Area of $△$CHD = $\frac{1}{2}$$\times$HD$\times$CH
= $\frac{1}{2}$$\times$7$\times$4 = $\frac{1}{2}$$\times$28 = 14 $c{m}^2$

Area of $△$BFA = $\frac{1}{2}$$\times$AF$\times$BF
= $\frac{1}{2}$$\times$6$\times$6 = $\frac{1}{2}$$\times$36 = 18 $c{m}^2$

area of $△$AED = $\frac{1}{2}$$\times$AD$\times$GE
= $\frac{1}{2}$$\times$(AF+FH+HD)$\times$5
= $\frac{1}{2}$$\times$(6+6+4)$\times$5
= $\frac{1}{2}$$\times$16$\times$5 = $\frac{1}{2}$$\times$80 = 40 $c{m}^2$

Area of given figure = 18+39+14+40
= 111 $c{m}^2$