Question

Find the area of letter E as shown in figure. The thickness is constant from all sides

• A. $10{\text{m}}^2$
• B. $13{\text{m}}^2$
• C. $14\text{m}$
• D. $21{\text{m}}^2$

Answer 1

The correct option is D $21{\text{m}}^2$

Lets divide the figure into different rectangles and name the vertices.
Area of a closed figure = Area of a rectangle AMKN + Area of a rectangle MBCD + Area of a rectangle IJNL + Area of reactangle EFGH

$\to$Area of rectangle AMKN
Given: The thickness is constant from all sides
$AM=1m(\because BD=FH=JL=1m)$

$\to$ Area of AMKN = length $\times$ breadth = AM $\times$ AK
$=1\times 8=8{\text{m}}^2$

$\to$ Area of MBCD = length $\times$ breadth = MB $\times$ MC
$=5\times 1=5{\text{m}}^2$

$\to$Area of IJNL = Area of MBCD = 5 m${}^2$

$\to$ Area of EFGH = length $\times$ breadth = EF $\times$ FH
$=3\times 1=3{\text{m}}^2$

Area of a closed figure = Area of a rectangle AMKN + Area of a rectangle MBCD + Area of a rectangle IJNL + Area of reactangle EFGH

Area of a closed figure $=8+5+5+3=21{\text{m}}^2$
Option (d) is correct.