Question-160839

Question Number 160839 by amin96 last updated on 07/Dec/21

Commented by amin96 last updated on 07/Dec/21

yellow area = ?

Answered by mr W last updated on 08/Dec/21

Commented by mr W last updated on 08/Dec/21

M e t h o d I (n o t s o s m a r t o n e)

\frac{A B}{\sin (α + 60)} = \frac{2}{\sin (α + 30)}

A B = \frac{2 \sin (α + 60)}{\sin (α + 30)}

\frac{D B}{\sin α} = \frac{2}{\sin (α + 30)}

D B = \frac{2 \sin α}{\sin (α + 30)}

S_{1} = \frac{1}{2} \times 2 \times \frac{2 \sin (α + 60)}{\sin (α + 30)} \times \sin 30 = \frac{\sin (α + 60)}{\sin (α + 30)}

S_{2} = \frac{1}{2} \times 2 \times \frac{2 \sin α}{\sin (α + 30)} \times \sin 30 = \frac{\sin α}{\sin (α + 30)}

S_{y e l l o w} = \frac{\sin (α + 60) + \sin α}{\sin (α + 30)}

= \frac{\sin (α + 30 + 30) + \sin (α + 30 - 30)}{\sin (α + 30)}

= \frac{2 \sin (α + 30) \cos 30}{\sin (α + 30)}

= 2 \cos 30 = \sqrt{3}

Commented by Tawa11 last updated on 08/Dec/21

Great sir .

Answered by mr W last updated on 08/Dec/21

Commented by mr W last updated on 08/Dec/21

M e t h o d I I

m a k e Δ D E A = Δ C D B a s s h o w n

y e l l o w a r e a = a r e a o f t r a p a z o i d A B D E

D E / / B A

B D = A E

B E = A D = 2

a r e a o f t r a p a z o i d = \frac{A D \times B E \times \sin 60}{2}

= 2 \sin 60 = \sqrt{3}

Commented by Ari last updated on 09/Dec/21

Mr.Why area of trapezoid is (AD*BE*sin60)/2.This fomule is true for each trapezoid?

Commented by mr W last updated on 09/Dec/21

{i t}^{'} s t r u e f o r a n y c o n v e x q u a d r i l a t e r a l!

A = \frac{d i a g o n a l 1 \times d i a g o n a l 2 \times \sin θ}{2}

Commented by Ari last updated on 09/Dec/21

ok,thanks!