In the above semicircle, chords A B , B C and C D have lengths 6 , 6 and 1 4 respectively. Find the area of quadrilateral A B C D to eight decimal places.
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Using the above diagram 2 ( 1 8 0 − 2 θ ) + λ = 1 8 0 ⟹ λ = 4 θ − 1 8 0 ⟹
m = 1 8 0 − 2 θ .
For △ A O B using the law of cosines we have:
3 6 = 2 r 2 ( 1 − cos ( 1 8 0 − 2 θ ) ) ⟹ 1 8 = r 2 ( 1 + cos ( 2 θ ) ) ⟹ r 2 = 1 + cos ( 2 θ ) 1 8
and
for △ C O D using the law of cosines we have:
1 9 6 − 2 r 2 ( 1 + cos ( 4 θ ) ) ⟹ 4 9 = r 2 ( cos 2 ( 2 θ ) ) ⟹
4 9 + 4 9 cos ( 2 θ ) = 1 8 cos 2 ( 2 θ ) ⟹ 1 8 cos 2 ( θ ) − 4 9 cos ( 2 θ ) − 4 9 = 0 ⟹
cos ( 2 θ ) = 2 4 9 ± 7 7 and ∣ cos ( u ) ∣ ≤ 1 ⟹ cos ( 2 θ ) = − 9 7
⟹ r 2 = 1 − 9 7 1 8 = 8 1 ⟹ r = 9 .
Let h 1 be the height of △ A O B ⟹ h 1 = 8 1 − 9 = 7 2 = 6 2 ⟹
A 1 = 2 ∗ A △ A O B = 3 6 2
Let h 2 be the height of △ C O D ⟹ h 2 = 8 1 − 4 9 = 3 2 = 4 2 ⟹
A 2 = A △ C O D = 2 8 2
⟹ Area of quadrilateral A B C D is A = A 1 + A 2 = 6 4 2 ≈ 9 0 . 5 0 9 6 6 7 9 9
@Rocco Dalto , you may specify 8 decimal places, but the system will just need 3 significant figure.
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Let ∠ C O D = α and ∠ C O B = ∠ B O A = β . Then α + 2 β = 1 8 0 ∘ . Let the radius of the semicircle be r . By cosine rule we have 2 r 2 ( 1 − cos α ) = 1 4 2 and 2 r 2 ( 1 − cos β ) = 6 2 . Therefore,
1 − cos β 1 − cos α 9 − 9 cos α − 9 cos ( 1 8 0 ∘ − 2 β ) + 4 9 cos β − 4 0 9 cos ( 2 β ) + 4 9 cos β − 4 0 1 8 cos 2 β + 4 9 cos β − 4 9 ⟹ cos β cos α = 6 2 1 4 2 = 4 9 − 4 9 cos β = 0 = 0 = 0 = 9 7 = − cos 2 β = sin 2 β − cos 2 β = − 8 1 1 7 Putting α = 1 8 0 ∘ − 2 β Note that cos ( 1 8 0 ∘ − θ ) = − cos θ and cos ( 2 θ ) = 2 cos 2 θ − 1 ⟹ sin β = 9 4 2 ⟹ sin α = 8 1 5 6 2
From 2 r 2 ( 1 − cos β ) = 6 2 ⟹ r 2 = 8 1 . Therefore, the area of the quadrilateral A = 2 1 r 2 sin α + r 2 sin β = 6 4 2 ≈ 9 0 . 5 0 9 6 6 7 9 9 .