In a figure, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D, E are the mid points of the sides AB and AC respectively, calculate the area of $Δ$ ADE

Question

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Answer 1

Given that $A B = 9 c m, A C = 15 c m$

$B C = A C^{2} - A B^{2}$

$B C = 1 5^{2} - 9^{2}$

$B C = 12 c m$

$D E$ = $\frac{B C}{2} = \frac{1 2}{2} = 6 c m$
D is the mid point of the side AB

AD = $\frac{A B}{2} = \frac{9}{2} = 4.5 c m$

$A r e a o f △ A D E = \frac{1}{2} \times A D \times D E = \frac{1}{2} \times 4.5 \times 6 = 13.5 c m^{2}$

Answer 2

Given that

A B = 9 c m, A C = 15 c m

B C = A C^{2} - A B^{2}

B C = 1 5^{2} - 9^{2}

B C = 12 c m

D E

=

\frac{B C}{2} = \frac{1 2}{2} = 6 c m

D is the mid point of the side AB

AD =

\frac{A B}{2} = \frac{9}{2} = 4.5 c m

A r e a o f △ A D E = \frac{1}{2} \times A D \times D E = \frac{1}{2} \times 4.5 \times 6 = 13.5 c m^{2}