Question

Activities-
(1) Calculate the area of the figures shown below: (2 Marks)

(2) Calculate the area of the figures shown below: (3 Marks)

Answer 1

(1) Consider the figure below.
Let's label the figure as follows:
In the figure, it is evident that $\angle ABC={90}^{\circ }$
Hence, AB perpendicular to BC.
The area of the right-angled triangle is half the product of the perpendicular sides. So, area of the right-angled triangle $=\frac{1}{2}\times AB\times BC$
$\begin{array}{cc}& \Rightarrow \frac{1}{2}\times 4.5\times 6\\ & \Rightarrow 13.5\text{Square centimeters}\end{array}$
(2) Consider the figure below.
Let's label the figure as follows:
In the figure, it is evident that $\angle ABC={90}^{\circ }$
Hence, $AB$ perpendicular to $BC$.
Here, ABED is a rectangle.
Area of the rectangle of length $l$ and breadth $b$ is $l\times b$
$l=6cm$ and $b=5cm$
So, the area of rectangle $ABED=6\times 5=30$ Square centimeters.
Consider $△ABC$
In the figure, it is evident that $\angle ABC={90}^{\circ }$
Hence, $AB$ perpendicular to $BC$.
Given that $BC=2cm$
Also, $BA=ED=5cm$ [Opposite sides of the rectangle are equal]
The area of the right-angled triangle is half the product of the perpendicular sides.
So, area of the right-angled triangle $=\frac{1}{2}\times BA\times BC$
$\Rightarrow \frac{1}{2}\times 2\times 5$
$\Rightarrow 5$ Square centimeters
Consider $△DEF$
Similarly, area of the right-angled triangle $=\frac{1}{2}\times ED\times DF$
Here, $ED=5cm$ and $DF=2cm$
$\begin{array}{cc}& \Rightarrow \frac{1}{2}\times 5\times 2\\ & \Rightarrow 5\text{Square centimeters}\end{array}$
So, total area $=$ Area of rectangle $ABCD+$ Area of $△ABC+$ Area of $△EDF$
$\Rightarrow$ Total area $=30+5+5$
$\Rightarrow$ Total area $=40$ Square centimeters