Question

Find the area of this figure.

• A. $30\text{m}$
• B. $30{\text{m}}^2$
• C. $15{\text{m}}^2$
• D. $60{\text{m}}^2$

Answer 1

The correct option is B $30{\text{m}}^2$

Let divide the figure into rectangles and triangles and name the vertices.
Area of closed figure = Area of a triangle ACD + Area of a rectangle ADBF + Area of a reactangle EFGH

Step 1: Find the area of a triangle ACD
$\text{Area of}△ABC=\frac{1}{2}\times \text{Base}\times \text{Height}$

To find base,
$CF=CE+EF=8+3=11m$ $\because EF=GH$
Base of a triangle $=CD=CF-AB=11-7=4m$

To find height,
Height of a triangle $=AD=BH-EG=6-4=2m$
Hence,
$\text{Area of}△ABC=\frac{1}{2}\times \text{Base}\times \text{Height}$
$\Rightarrow \frac{1}{2}\times 4\times 2=4m^2$

Step 2: Now, find the area of a rectangle ADBF
$\to$ Area of a rectangle ADBF = length $\times$ breadth = AB $\times$ AD
$=7\times 2=14{\text{m}}^2(\because AD=2m)$

Step 3: Now, find the area of a rectangle EFGH
$\to$ Area of a rectangle EFGH = length $\times$ breadth = GH $\times$ EG
$=3\times 4=12{\text{m}}^2$

Step 4: Now, find the area of a closed figure
Area of a closed figure $=4m^2+14m^2+12m^2=30{\text{m}}^2$
Option (b) is correct.