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In given figu...

Question

In given figure, $A B C D$ is a parallelogram, $A B = 30 c m, A D = 20 c m, D E = 10 c m$ and $C N = 27 c m$ , then the ratio of the area of ABCD to the area of $Δ A D E$ is,

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Solution

$l e t ∠ A D E = θ$

$a r e a o f p a r a l l e l o g r a m$

$A B C D = 20 \times 30 s i n θ = 600 s i n θ$

$(∵ θ i s a n g l e b e t w e e n a d j a c e n t s i d e s)$

$A r e a o f △ A D E = \frac{1}{2} \times 20 \times 10 \times s i n θ = 100 s i n θ$

$\frac{A r e a o f A B C D}{A r e a o f A D E} = \frac{600 s i n θ}{100 s i n θ} = 6$

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