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Math Help - Area of triangle

  1. #1
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    Area of triangle

    Area of triangle-pic2.jpg
    Given PR = 20 , QS = 15 , PR = x , QS = y , angle PTQ = 60 degree

    (a)
    Find the area of triangle PQT

    (b)
    Find the area of triangle QRT

    (c)
    Using (a) and (b), find the area of PQRS

    -------------------------------------------
    (a)
    triangle PQT =  \frac {\sqrt 3}{ 4 }xy
    (b)
    triangle QRT = \frac{\sqrt 3}{4}(20-x)y<br />
    (c)
    don't know how to do (c)

    Thanks in advance!
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  2. #2
    MHF Contributor
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    Hello cakeboby
    Quote Originally Posted by cakeboby View Post
    Click image for larger version. 

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    Given PR = 20 , QS = 15 , PR = x , QS = y , angle PTQ = 60 degree

    (a)
    Find the area of triangle PQT

    (b)
    Find the area of triangle QRT

    (c)
    Using (a) and (b), find the area of PQRS

    -------------------------------------------
    (a)
    triangle PQT =  \frac {\sqrt 3}{ 4 }xy
    (b)
    triangle QRT = \frac{\sqrt 3}{4}(20-x)y<br />
    (c)
    don't know how to do (c)

    Thanks in advance!
    You've clearly made a couple of typos in the 'given', but judging by the answers to (a) and (b), I assume you mean
    PT = x and QT = y
    and you've then used the \tfrac12ab\sin C formula for the area of a triangle, where \sin 60^o =\frac{\sqrt3}{2}.

    For (c), use the same method again to find the areas of \triangle s PTS and STR (where TS = 15-y), and then add all four areas together.

    Can you complete it now?

    Grandad
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