Question
Construct a rectangle $\text{PQRS}$, when its Area $=21 \ cm^2$ and length $=4.2 \ cm$
✓
Answer
Since area of rectangle $=21 \ cm ^2$
And, length $=4.2 \ cm$
Breadth $=$ Area $\div$ Length $=21 \div 4.2=5 \ cm$
Steps of construction:
$1$. Draw $B C=5 \ cm$
$2$. Through $B$, draw $B P$ such that $\angle B=90^{\circ}$
$3$. From $B P$, cut $B A=4.2 \ cm$
$4.$ With $A$ and $C$ as centres and radii $5 \ cm$ and $4.2 \ cm$ respectively, draw arcs cutting each other at $D$.
$5$. Join $A D$ and $C D$.
Thus, $\text{ABCD}$ is the required triangle.
And, length $=4.2 \ cm$
Breadth $=$ Area $\div$ Length $=21 \div 4.2=5 \ cm$
Steps of construction:
$1$. Draw $B C=5 \ cm$
$2$. Through $B$, draw $B P$ such that $\angle B=90^{\circ}$
$3$. From $B P$, cut $B A=4.2 \ cm$
$4.$ With $A$ and $C$ as centres and radii $5 \ cm$ and $4.2 \ cm$ respectively, draw arcs cutting each other at $D$.
$5$. Join $A D$ and $C D$.
Thus, $\text{ABCD}$ is the required triangle.
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