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    (Original post by atsruser)
    What's your solution?
    Post 3 + Post 11

    2 right angled triangles followed by similar triangles
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    It's not a module question so surely somebody can post a full answer?
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    I think the answer 48. and for more you share your solution.
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    (Original post by The Big Stone)

    i am stuck on this question, not really sure how to approach it. i tried to find the height of the triangle, but couldn't work it out because i had too many unknowns. i can't figure out a way to use the info to find the answer.

    the answer, for those who are wondering is D
    Another approach:

    Draw segment TS.

    Note that QRST is an orthogonal quadrilateral (diagonals are perpendicular).Note that with all orthogonal quadrilaterals, you can reflect the 4 triangles across the respective hypotenuse of each, creating a rectangle with double the area and dimensions equal to the diagonal lengths.

    Note that there's an SAS similarity, and that the ratio of areas of similar polygons is the square of the ratio of their side-lengths.
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    (Original post by MathMeister)
    It's not a module question so surely somebody can post a full answer?
    Method 1: (suggested by me, improved by The Big Stone)

    a) Triangles QTR and TPR have the same base and height, and hence have the same area. Thus area QPR = 2 x area QTR.

    b) Let the medians intersect at O. The medians of a triangle divide each other in the ratio 2:1 and since QS = 8, then QO = 16/3, OS = 8/3.

    c) Hence area QTR = 1/2 x 12 x 16/3 = 32 and so area QPR = 2 x 32 = 64.

    Method 2: (suggested by TenOfThem and aznkid66)

    a) Construct TS. Then area QTSR = 1/2 x 8 x 12 = 48. (Why? - construct a suitable rectangle if this is not clear)

    b) Note that PT=TQ and PS=SR, hence PTS and PQR are similar with scale factor 2, since angle QPR is common to both and PT:PQ = PS:PR = 1:2.

    So:

    area PQR = 4 x area PTS
    area PTS = area PQR - 48

    giving:

    area PQR = 4(area PQR - 48) = 4 x area PQS - 192

    which solves to give area PQR = 64

    Method 3:

    I've also seen another approach based on the fact that:

    area QSR: area OSR = QS : QO

    This comes from some triangle theorem whose name I don't know.
 
 
 
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