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# help

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A semicircle is inscribed in a quadrilateral ABCD in such a way that the midpoint of BC coincides with the center of the semicircle, and the diameter of the semicircle lies along a portion of BC. If AB = 4 and CD = 5, what is BC?

Nov 26, 2019

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Is there a diagram?

Nov 26, 2019
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A semicircle is inscribed in a quadrilateral ABCD in such a way that the midpoint of BC coincides with the center of the semicircle, and the diameter of the semicircle lies along a portion of BC. If AB = 4 and CD = 5, what is BC?

I have not solved this problem.

I have only drawn an initial diagram.

Here it is.  W is the centre of the circle and also the centre of side BC of the quadrilateral Nov 27, 2019
edited by Melody  Nov 27, 2019
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Maybe you need to use some obscure circle geomtery ratio  idk

It is driving me nuts. Can someone please show me how do do it...

Nov 27, 2019
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Deleted. Just noticed my semicircles weren't fully inscribed!!

Alan  Nov 28, 2019
edited by Alan  Nov 28, 2019
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I get an answer of 4sqrt(5).

I need to check my algebra and, if correct, I'll try to a post a solution tomorrow, (no time at the moment).

Nov 28, 2019
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Thanks Tiggsy Melody  Nov 28, 2019
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+1 Nov 29, 2019
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See diagram from previous post.

Triangles BEO and CGO are congruent.

Let BE = x , the angle at E is a right angle so  x = (R + h)cos(theta).

EA = 5 - x, and if AD is to be a tangent to the semi circle, then the line from A to the point of tangency will have length 5 - x also.

Similarly on the RHS, the line from D to the point of tangency will have length 4 - x.

So, if AD is to be a tangent to the semi circle, it will have a length 9 - 2x = 9 - 2(R + h)cos(theta).

Now move into co-ordinate geometry mode.

Take BC to be the horizontal axis, with O as the origin.

Point A will have an x co-ordinate   -(R + h) + 5cos(theta) and a y co-ordinate 5sin(theta).

Point D will have an x co-ordinate    (R + h) - 4cos(theta) and a y co-ordinate 4sin(theta).

So,

$$\displaystyle AD^{2}= \{9-2(R+h)\cos\theta\}^{2}=\{(R+h)-4\cos\theta-\{-(R+h)+5\cos\theta\}\}^{2}+(4\sin\theta-5\sin\theta)^{2},\\ 81-36(R+h)\cos\theta+4(R+h)^{2}\cos^{2}\theta=4(R+h)^{2}-36(R+h)\cos\theta+81\cos^{2}\theta+\sin^{2}\theta,\\ 81+4(R+h)^{2}\cos^{2}\theta-4(R+h)^{2}-81\cos^{2}\theta-\sin^{2}\theta = 0, \\ 4(R+h)^{2}(\cos^{2}\theta-1)+81\sin^{2}\theta -\sin^{2}\theta=0,\\ 4(R+h)^{2}(-\sin^{2}\theta)+80\sin^{2}\theta=0,\\ (R+h)^{2}=20,\\ R+h =\sqrt{20}=2\sqrt{5} .$$

$$\displaystyle \text{BC}=2(R+h)=4\sqrt{5}.$$

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Nov 29, 2019
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Thanks very much Tiggsy.  I have not studied it yet but I will. Nov 29, 2019