Question-183886

Question Number 183886 by Acem last updated on 31/Dec/22

Answered by HeferH last updated on 31/Dec/22

A = (\frac{B C + A D}{2}) 15

A E = \sqrt{17^{2} - 15^{2}} = \sqrt{2 \cdot 32} = 8

B^{'} D = \sqrt{39^{2} - 15^{2}} = 36

A D = A E + E D = 8 + 36 - B C \Rightarrow

A = (\frac{8 + 36}{2}) 15 = 66 \cdot 5 = 330 u

Commented by Acem last updated on 31/Dec/22

V e r y w e l l S i r!

Answered by mr W last updated on 31/Dec/22

Commented by mr W last updated on 31/Dec/22

[A B C D] = [Δ A B F] = \frac{A F \times B G}{2}

= \frac{(\sqrt{17^{2} - 15^{2}} + \sqrt{39^{2} - 15^{2}}) \times 15}{2}

= \frac{44 \times 15}{2} = 330 ✓

Commented by manxsol last updated on 31/Dec/22

t h e f o r m u l a w o u l d t h e n b e

A r e a t r a p (d 1, d 2, h) =

[\sqrt{d 1^{2} - h^{2}} + \sqrt{d 2^{2} - h^{2}]} \times \frac{h}{2}

Commented by manxsol last updated on 31/Dec/22

Commented by Acem last updated on 31/Dec/22

V e r y w e l l S i r!

Answered by mr W last updated on 31/Dec/22

a l t e r n a t i v e :

\cos α = \frac{15}{17} \Rightarrow \sin α = \frac{\sqrt{17^{2} - 15^{2}}}{17} = \frac{8}{17}

\cos β = \frac{15}{39} \Rightarrow \sin α = \frac{\sqrt{39^{2} - 15^{2}}}{39} = \frac{36}{39}

[A B C D] = \frac{1}{2} \times 17 \times 39 \times \sin (α + β)

= \frac{1}{2} \times 17 \times 39 \times (\frac{8}{17} \times \frac{15}{39} + \frac{15}{17} \times \frac{36}{39})

= \frac{15 \times 44}{2} = 330 ✓

Commented by Acem last updated on 31/Dec/22

V e r y w e l l S i r!

a n d p l e a s e e x p l a n a t i o n a b o u t \sin (α + β) t h a n k s

Commented by mr W last updated on 31/Dec/22

Commented by mr W last updated on 31/Dec/22

A = \frac{1}{2} d_{1} \times d_{2} \times \sin θ

θ = α + β

Commented by Acem last updated on 31/Dec/22

E x c e l l e n t! T h a n k y o u d e a r f r i e n d

Answered by Acem last updated on 31/Dec/22

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