Question 11.

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

A
B
C
D

Question Explanation

Text Explanation


Now area of ADE=12×AD×AE×sinAADE = \frac{1}{2} \times AD \times AE \times \sin A

=12×2x×2y×sinA=8= \frac{1}{2} \times 2x \times 2y \times \sin A = 8

we get  

xysinA=4xy \sin A = 4

Now area of triangle ABC=12×AB×AC×sinAABC = \frac{1}{2} \times AB \times AC \times \sin A

we get  

12×3x×5y×sinA=152xysinA=152×4\frac{1}{2} \times 3x \times 5y \times \sin A = \frac{15}{2} xy \sin A = \frac{15}{2} \times 4

We get area of ABC = 30

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