Chords Of A Semicircle

Geometry Level 3

In the above semicircle, chords A B , B C AB, BC and C D CD have lengths 6 , 6 6,6 and 14 14 respectively. Find the area of quadrilateral A B C D ABCD to eight decimal places.


The answer is 90.50966799.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Chew-Seong Cheong
Jan 13, 2020

Let C O D = α \angle COD = \alpha and C O B = B O A = β \angle COB = \angle BOA = \beta . Then α + 2 β = 18 0 \alpha + 2\beta = 180^\circ . Let the radius of the semicircle be r r . By cosine rule we have 2 r 2 ( 1 cos α ) = 1 4 2 2r^2(1-\cos \alpha) = 14^2 and 2 r 2 ( 1 cos β ) = 6 2 2r^2 (1-\cos \beta) = 6^2 . Therefore,

1 cos α 1 cos β = 1 4 2 6 2 9 9 cos α = 49 49 cos β Putting α = 18 0 2 β 9 cos ( 18 0 2 β ) + 49 cos β 40 = 0 Note that cos ( 18 0 θ ) = cos θ 9 cos ( 2 β ) + 49 cos β 40 = 0 and cos ( 2 θ ) = 2 cos 2 θ 1 18 cos 2 β + 49 cos β 49 = 0 cos β = 7 9 sin β = 4 2 9 cos α = cos 2 β = sin 2 β cos 2 β = 17 81 sin α = 56 2 81 \begin{aligned} \frac {1-\cos \alpha}{1-\cos \beta} & = \frac {14^2}{6^2} \\ 9 - 9 \cos \alpha & = 49 - 49 \cos \beta & \small \blue{\text{Putting }\alpha = 180^\circ - 2\beta} \\ - 9 \cos (180^\circ - 2\beta) + 49 \cos \beta - 40 & = 0 & \small \blue{\text{Note that }\cos (180^\circ -\theta)= - \cos \theta} \\ 9 \cos (2\beta) + 49 \cos \beta - 40 & = 0 & \small \blue{\text{and }\cos (2\theta)= 2\cos^2 \theta-1} \\ 18\cos^2 \beta + 49 \cos \beta - 49 & = 0 \\ \implies \cos \beta & = \frac 79 & \small \blue{\implies \sin \beta = \frac {4\sqrt 2}9} \\ \cos \alpha & = - \cos 2 \beta \\ & = \sin^2 \beta - \cos^2 \beta \\ & = - \frac {17}{81} & \small \blue{\implies \sin \alpha = \frac {56\sqrt 2}{81}} \end{aligned}

From 2 r 2 ( 1 cos β ) = 6 2 r 2 = 81 2r^2(1-\cos \beta) = 6^2 \implies r^2 = 81 . Therefore, the area of the quadrilateral A = 1 2 r 2 sin α + r 2 sin β = 64 2 90.50966799 A = \dfrac 12r^2 \sin \alpha + r^2 \sin \beta = 64\sqrt 2 \approx \boxed{90.50966799} .

Rocco Dalto
Jan 12, 2020

Using the above diagram 2 ( 180 2 θ ) + λ = 180 λ = 4 θ 180 2(180 - 2\theta) + \lambda = 180 \implies \lambda = 4\theta - 180 \implies

m = 180 2 θ m = 180 - 2\theta .

For A O B \triangle{AOB} using the law of cosines we have:

36 = 2 r 2 ( 1 cos ( 180 2 θ ) ) 18 = r 2 ( 1 + cos ( 2 θ ) ) r 2 = 18 1 + cos ( 2 θ ) 36 = 2r^2(1 - \cos(180 - 2\theta)) \implies 18 = r^2(1 + \cos(2\theta)) \implies r^2 = \dfrac{18}{1 + \cos(2\theta)}

and

for C O D \triangle{COD} using the law of cosines we have:

196 2 r 2 ( 1 + cos ( 4 θ ) ) 49 = r 2 ( cos 2 ( 2 θ ) ) 196 - 2r^2(1 + \cos(4\theta)) \implies 49 = r^2(\cos^2(2\theta)) \implies

49 + 49 cos ( 2 θ ) = 18 cos 2 ( 2 θ ) 18 cos 2 ( θ ) 49 cos ( 2 θ ) 49 = 0 49 + 49\cos(2\theta) = 18\cos^2(2\theta) \implies 18\cos^2(\theta) - 49\cos(2\theta) - 49 = 0 \implies

cos ( 2 θ ) = 49 ± 77 2 \cos(2\theta) = \dfrac{49 \pm 77}{2} and cos ( u ) 1 cos ( 2 θ ) = 7 9 |\cos(u)| \leq 1 \implies \cos(2\theta) = -\dfrac{7}{9}

r 2 = 18 1 7 9 = 81 r = 9. \implies r^2 = \dfrac{18}{1 - \dfrac{7}{9}} = 81 \implies r = 9.

Let h 1 h_{1} be the height of A O B h 1 = 81 9 = 72 = 6 2 \triangle{AOB} \implies h_{1} = \sqrt{81 - 9} = \sqrt{72} = 6\sqrt{2} \implies

A 1 = 2 A A O B = 36 2 A_{1} = 2*A_{\triangle{AOB}} = 36\sqrt{2}

Let h 2 h_{2} be the height of C O D h 2 = 81 49 = 32 = 4 2 \triangle{COD} \implies h_{2} = \sqrt{81 -49} = \sqrt{32} = 4\sqrt{2} \implies

A 2 = A C O D = 28 2 A_{2} = A_{\triangle{COD}} = 28\sqrt{2}

\implies Area of quadrilateral A B C D ABCD is A = A 1 + A 2 = 64 2 90.50966799 A = A_{1} + A_{2} = 64\sqrt{2} \approx \boxed{90.50966799}

@Rocco Dalto , you may specify 8 decimal places, but the system will just need 3 significant figure.

Chew-Seong Cheong - 1 year, 5 months ago

Log in to reply

I wasn't aware of that, Thanks.

Rocco Dalto - 1 year, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...