In quadrilateral A B C D above, ∠ B A C = 3 x , ∠ B C A = x , ∠ C A D = 5 x , ∠ A C D = 3 x and B C = C D = 1 . Find the area of quadrilateral A B C D .
Note: A variation of the above problem was taken from a problem I found online.
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... = 4 3
which is the area of an equilateral triangle with side 1 . I wonder if there's a direct geometric way to show that?
There is an excellent geometric proof of the fact that x = 1 0 ∘ by @Maria Kozlowska here .
From the angle conditions (solutions above showed that x = 1 0 ∘ ), we know that A B ∣ ∣ D C and so the quadrilateral is a trapezium, the height of the trapezium is 1 × sin ( 1 8 0 ∘ − ∠ A B C ) = sin 4 0 ∘ .
By sine rule, A B = sin 1 0 ∘ × sin 3 0 ∘ 1 = 2 sin 1 0 ∘ . So the area of quadrilateral A B C D is 2 1 ( 1 + 2 sin 1 0 ∘ ) ( sin 4 0 ∘ ) ≈ 0 . 4 3 3 0 1 2 7 0
By sine rule :
B C A C sin 3 x sin ( 1 8 0 ∘ − 4 x ) sin 3 x sin 4 x sin 3 x 1 sin 3 x sin 5 x sin 3 x sin 5 x ⟹ 5 x ⟹ 3 x = C D A C = 1 A C = sin 5 x sin ( 1 8 0 ∘ − 8 x ) = sin 5 x sin 8 x = sin 5 x 2 cos 4 x = 2 cos 4 x = 2 1 sin ( 9 0 ∘ − 4 x ) = 9 0 ∘ − 4 x ⟹ x = 1 0 ∘ = 3 0 ∘ ⟹ x = 1 0 ∘ SInce sin ( 1 8 0 ∘ − θ ) = sin θ And sin 2 θ = 2 sin θ cos θ Also sin θ = cos ( 9 0 ∘ − θ )
Fill in the measures of all the angles. Extend B C and D A to meet at E . Then we note that ∠ E = 4 0 ∘ and △ C D E and △ A B E are isosceles and similar. Therefore the area of quadrilateral A B C D :
[ A B C D ] = [ C D E ] − [ A B E ] = 2 1 ⋅ 1 2 ⋅ sin 8 0 ∘ − 2 1 ⋅ A B 2 ⋅ sin 8 0 ∘ = 2 sin 8 0 ∘ ( 1 − sin 2 3 0 ∘ sin 2 1 0 ∘ ) ≈ 0 . 4 3 3 By sine rule, A B = sin 3 0 ∘ sin 1 0 ∘
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For △ A D C :
D C sin ( 5 x ) = A C sin ( 8 x ) ⟹ sin ( 8 x ) sin ( 5 x ) = A C D C = A C B C = A C 1
For △ A B C :
B C sin ( 3 x ) = A C sin ( 4 x ) ⟹ sin ( 4 x ) sin ( 3 x ) = A C B C = = A C 1
⟹
sin ( 8 x ) sin ( 5 x ) = sin ( 4 x ) sin ( 3 x ) ⟹ sin ( 5 x ) sin ( 4 x ) = sin ( 8 x ) sin ( 3 x ) =
2 sin ( 4 x ) cos ( 4 x ) sin ( 3 x ) ⟹ sin ( 5 x ) = 2 cos ( 4 x ) sin ( 3 x )
Using the identity 2 cos ( A ) sin ( B ) = sin ( A + B ) − sin ( A − B ) , with A = 4 x and
B = 3 x we obtain: sin ( 5 x ) = sin ( 7 x ) − sin ( x ) ⟹ sin ( 7 x ) − sin ( 5 x ) = sin ( x )
or sin ( 7 x ) + sin ( − 5 x ) = sin ( x )
Using the identity sin ( A ) + sin ( B ) = 2 sin ( 2 A + B ) cos ( 2 A − B ) , with A = 7 x and
B = − 5 x we obtain: 2 sin ( x ) cos ( 6 x ) = sin ( x ) ⟹ 2 cos ( 6 x ) = 1 ⟹ cos ( 6 x ) = 2 1
⟹ 6 x = 6 0 ∘ ⟹ x = 1 0 ∘
Using the above diagram with x = 1 0 ∘ ⟹ h 2 = sin ( 3 0 ∘ ) = 2 1 and h 1 = sin ( 1 0 ∘ )
For A C we have: sin ( 1 4 0 ∘ ) A C = sin ( 3 0 ∘ ) 1 = 2 ⟹ A C = 2 sin ( 1 4 0 ∘ )
⟹ The total area A A B C D = 2 1 ( 2 sin ( 1 4 0 ∘ ) sin ( 1 0 ∘ ) ) + 2 1 ( 2 sin ( 1 4 0 ∘ ) ) ( 2 1 ) =
sin ( 1 4 0 ∘ ) ( sin ( 1 0 ∘ ) + 2 1 ) = 2 1 sin ( 1 4 0 ∘ ) ( 2 sin ( 1 0 ∘ ) + 1 ) ≈ 0 . 4 3 3 0 1 2 7 0