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Thread: [ASK] Question About Circle and Parallelogram

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    [ASK] Question About Circle and Parallelogram

    A parallelogram ABCD has angle A = angle C = 45. Circle K with the center C intercept the parallelogram through B and D. AD is extended so that it intercepts the circle at E and BE intercepts CD at H. The ratio of the area of triangle BCH and triangle EHD is ....

    Here I got that triangle BCH and triangle EHD is similar with angle BCH = angle HDE = 45, angle CHB = angle DHE = 112.5, and angle CBH = angle DEH = 22.5. The area of triangle BCH is CH h, where h is the parallelogram's height. The area of triangle EHD is (r - CH) r. I stuck at the ratio is (CH h) : ((r - CH) r). How do I simplify that? Problem is, I don't know CD's length, so I can't approximate the ratio between CH and DH.
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    Re: [ASK] Question About Circle and Parallelogram

    Draw your parallelogram with AB horizontal and labelled counterclockwise ABCD. Since CD = CB = r, the radius of the circle, your parallelogram is a rhombus of side r. Draw line CE and Notice that triangle CDE is isosceles with two equal sides r, angle CDE is $45^\circ$ and DCE is a right angle. So DE is $r\sqrt 2$ and BC = r. Areas are proportional to the square of the similar sides.
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    Re: [ASK] Question About Circle and Parallelogram

    Quote Originally Posted by Walagaster View Post
    Since CD = CB = r, the radius of the circle, your parallelogram is a rhombus of side r.
    Why did I miss this...

    Quote Originally Posted by Walagaster View Post
    Areas are proportional to the square of the similar sides.
    How do you get this?
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    Re: [ASK] Question About Circle and Parallelogram

    Quote Originally Posted by Monoxdifly View Post
    Why did I miss this...


    How do you get this?
    It's true for general planar areas. It's easy enough to check for triangles. Suppose you have two similar triangles where the linear dimensions (including base and height) of one are $k$ times the corresponding dimensions of the other. You can check for yourself how their areas compare. What is the ratio of the areas?
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    Re: [ASK] Question About Circle and Parallelogram

    Quote Originally Posted by Walagaster View Post
    It's true for general planar areas. It's easy enough to check for triangles. Suppose you have two similar triangles where the linear dimensions (including base and height) of one are $k$ times the corresponding dimensions of the other. You can check for yourself how their areas compare. What is the ratio of the areas?
    Let's see...
    $\displaystyle (\frac12rh) : (\frac12r\sqrt2kh)$
    Wait, are you implying that $\displaystyle k=\sqrt2$?
    Last edited by Monoxdifly; Jul 7th 2019 at 05:11 PM.
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    Re: [ASK] Question About Circle and Parallelogram

    Quote Originally Posted by Walagaster
    Areas are proportional to the square of the similar sides.
    Quote Originally Posted by Monoxdifly
    How do you get this?
    Quote Originally Posted by Walagaster
    It's true for general planar areas. It's easy enough to check for triangles. Suppose you have two similar triangles where the linear dimensions (including base and height) of one are k times the corresponding dimensions of the other. You can check for yourself how their areas compare. What is the ratio of the areas?
    Quote Originally Posted by Monoxdifly View Post
    Let's see...
    $\displaystyle (\frac12rh) : (\frac12r\sqrt2kh)$
    Wait, are you implying that $\displaystyle k=\sqrt2$?
    Please don't mix up two posts. I said areas are proportional to the square of similar sides. You asked how I got that. I explained it in the last post above. Then I asked you to check it out. Please reply to the blue question so you understand it before trying to apply it to your problem.
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    Re: [ASK] Question About Circle and Parallelogram

    From the ratio itself I got $\displaystyle k=\sqrt2$. Thus, the ratio itself is 1 : 2.
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    Re: [ASK] Question About Circle and Parallelogram

    Quote Originally Posted by Monoxdifly View Post
    From the ratio itself I got $\displaystyle k=\sqrt2$. Thus, the ratio itself is 1 : 2.
    You need to learn to reply with quote and address the issues in what you are quoting. I gather that you have figured out that the area of one triangle in the original problem is twice the area of the other. But your communication skills leave a lot to be desired and I can't tell if you have really understood what I have told you. So good luck and I'm done with this thread.
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    Re: [ASK] Question About Circle and Parallelogram

    Quote Originally Posted by Walagaster View Post
    You need to learn to reply with quote and address the issues in what you are quoting. I gather that you have figured out that the area of one triangle in the original problem is twice the area of the other. But your communication skills leave a lot to be desired and I can't tell if you have really understood what I have told you. So good luck and I'm done with this thread.
    Even though I misunderstood what you meant, thanks to your hint I could figure out the answer though we might reach it from different paths. But since you said you're done with this thread, well, thanks for your help.
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