Question

A big triangular field is divided into two parts as shown in the figure below. What is the area of smaller triangular field ?

Given- AB = 8 m, AC = 10 m, DE = 3 m, $\angle ABC$ = $\angle AED$ = ${90}^{\circ }$

• A. 12 $m^2$
• B. 8 $m^2$
• C. 10 $m^2$
• D. 6 $m^2$

Answer 1

The correct option is D

6 $m^2$

In right angled triangle ABC,

$A{B}^2$+$B{C}^2$ = $A{C}^2$ (Pythagoras theorem)

$B{C}^2$ = ${10}^2$-$8^2$

$B{C}^2$ = 100-64=36

BC = 6 m

Area of $△ABC$= $\frac{1}{2}\times base\times height$

= $\frac{1}{2}\times BC\times AB$

= $\frac{1}{2}\times 6\times 8$

= 24 $m^2$

In $△ABC$ and $△ADE$,

$\angle ABC$ = $\angle AED$ = ${90}^{\circ }$

and $\angle A$ is common to both.

$\therefore$ By A.A. axiom, $△ABC$ is similar to $△ADE$

$\frac{Areaof△ADE}{Areaof△ABC}$ = $\frac{D{E}^2}{B{C}^2}$

$\frac{Areaof△ADE}{24m^2}$ = $\frac{3^2}{6^2}$

$Areaof△ADE$ = $\frac{9}{36}\times 24$ $m^2$

= 6 $m^2$