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Math Help - Question on Similarity

  1. #1
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    Exclamation Question on Similarity

    Doc1.doc
    The diagram is attached.
    The question is:
    In the diagram ABCD is a parallelogram. If BC= 4cm, CP= 6 cm and the area of ABCD = 40 cm^2 , find the area of triangle ABP.
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  2. #2
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    The area of a parallelogram is "height times base". Since you are told that the base (the length of AD which is the same as the length of BC) is 4 and that the area of the parallelogram is 40, the height, the perpendicular distance from BC to AD, is 10.

    The area of a triangle is "(1/2) heigth times base" and the base of triangle APD is also 4 so you need to find the perpendicular distance from P to AD. That will be 10 plus the perpendicular distance from P to BC. The two right triangles having hypotenuses CP and DC are similar triangles but, since you do not know CD, I see no way of using that to get the perpendicular distance you need.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    The area of a parallelogram is "height times base". Since you are told that the base (the length of AD which is the same as the length of BC) is 4 and that the area of the parallelogram is 40, the height, the perpendicular distance from BC to AD, is 10.

    The area of a triangle is "(1/2) heigth times base" and the base of triangle APD is also 4 so you need to find the perpendicular distance from P to AD. That will be 10 plus the perpendicular distance from P to BC. The two right triangles having hypotenuses CP and DC are similar triangles but, since you do not know CD, I see no way of using that to get the perpendicular distance you need.
    To get a triangle as APD, a new line segment will have to be made. Can you please explain, 'the perpendicular distance from P to AD. That will be 10 plus the perpendicular distance from P to BC.' ?
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  4. #4
    Senior Member Sambit's Avatar
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    I too tried a lot to get something helpful...but did not get any. There is possibly no way to answer your question...
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  5. #5
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    Quote Originally Posted by Ilsa View Post
    To get a triangle as APD, a new line segment will have to be made. Can you please explain, 'the perpendicular distance from P to AD. That will be 10 plus the perpendicular distance from P to BC.' ?
    Yes, and that new line segment is the line from P perendicular to line AD. The length of that line segment is 'the perpendicular distance from P to AD'. The length of the line segment from P to perpendicular BC is the second "perpendicular distance".
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