Question

If the areas of triangles APB and PCB are $60{\text{cm}}^2$ and $40{\text{cm}}^2$ respectively and if their heights are 6 cm and 5 cm respectively, then what is the perimeter of the rectangle ABCD?

• A. $18\text{cm}$
• B. $36\text{cm}$
• C. $72\text{cm}$
• D. $80\text{cm}$

Answer 1

The correct option is C $72\text{cm}$

From the diagram it is clear that,
base of $△APB$ = length of the rectangle ABCD
base of $△PCB$= breadth of the rectangle ABCD

Area of a triangle = $\frac{1}{2}$$\times$base$\times$height
$⟹\text{Area of}\Delta APB=60=\frac{1}{2}\times AB\times 6$
$⟹AB=20cm$
Similarly, $\text{Area of}\Delta PBC=40=\frac{1}{2}\times BC\times 5$
$⟹BC=16cm$
Now, $\text{perimeter of a rectangle}=2\times \left(\text{length+breadth}\right)$
= 2(20+16)
= $72\text{cm}$