Question

Find the area of the figure.

Answer 1

Let us split the given figure into the following individual regions.

• Rectangle: HBDG
• Right-angled triangles: ABI, EDF
• Acute-angled triangle: BCD

Area of the rectangle HBDG
= Length $\times$ Width
= 9 in $\times$ 3 in
= 27 in${}^2$

Area of $△$ABI
$=\frac{1}{2}\times \text{Base}\times \text{Height}=\frac{1}{2}\times BI\times AI=\frac{1}{2}\times (9-6)\times 4=2\times 3=6{\text{in}}^2$

Similarly, area of $△$EDF = 6 in${}^2$

Area of $△$BCD
$=\frac{1}{2}\times \text{Base}\times \text{Height}=\frac{1}{2}\times BD\times CJ=\frac{1}{2}\times HG\times CJ=\frac{1}{2}\times 3\times 5=\frac{15}{2}{\text{in}}^2$

$\therefore$ Area of the given figure
$=$ Sum of the areas of individual regions
$=27+6+6+\frac{15}{2}=39+\frac{15}{2}=\frac{78+15}{2}=\frac{93}{2}i{n}^2$