Question

A rectangle of area $48c{m}^2$ is divided into two congruent triangles across its diagonal. Find the sum of the areas of both the triangles.

• A. $24c{m}^2$
• B. $48c{m}^2$
• C. $28c{m}^2$
• D. $30c{m}^2$

Answer 1

The correct option is B

$48c{m}^2$

Area of rectangle = $length\times breadth$
We get two congruent triangles, when a rectangle is cut across its diagonal.
$\therefore$ Area of the rectangle = Sum of the areas of the two congruent triangles.
As the triangles are congruent, their areas are equal.
Hence,Area of each triangle = $\frac{Areaoftherectangle}{2}$
$\frac{48c{m}^2}{2}=24c{m}^2$
Thus the area of one of the triangles = $24c{m}^2$
The 2nd triangle will also have an area of $24c{m}^2$, as it is congruent to the first.
$\therefore$ Sum of the areas of both the triangles
= $24c{m}^2$ + $24c{m}^2$ = $48c{m}^2$