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.

Answer 2

From the above, we know that
Area of △ADE = \(\frac{AD}{AB}\times\frac{AE}{AC}\times\text{Area of △ABC}\)
⇒ \(8=\frac{2}{3}\times\frac{2}{5}\times\text{Area of △ABC}\)
⇒ Area of △ABC = \(\frac{8\times3\times5}{2\times2}\)
= \(\frac{120}{4}\)
= 30
So , the Area of △ABC is 30 cm²
Therefore, the correct answer is 30.