Question

Let D and E be points on sides AB and AC, respectively, of a triangle ABC, such that AD : BD = 2 : 1 and AE : CE = 2 : 3. If the area of the triangle ADE is 8 sq cm, then the area of the triangle ABC, in sq cm, is

Answer 1

Area of $ΔADE=\frac{1}{2}\times AD\times AE\times sinA$
$=\frac{1}{2}\times 2x\times 2y\times SinA=8$
$⇒ xy SinA=4$
The area of triangle ABC is now calculated as $\frac{1}{2}\times AB\times A\times sinA$
$=\frac{1}{2}\times 3x\times 5y\times sinA$
$⇒\frac{15}{2}xy\space sinA=\frac{15}{4}\times4=30$
$∴$ Area of $ABC = 30$
Therefore, the correct answer is 30 cm.