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Math Help - Desperately need help!

  1. #1
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    Desperately need help!

    PQR is a right angled triangle with PR = 3cm and QR = 4cm. The square STUV is inscribed in triangle PQR. What is the length, in centimeters of the side of the square.

    A) 30/17
    B) 12/7
    C) 5/3
    D) 60/37
    E) 60/39

    Please help!! and tell me how you did it..
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by jgv115 View Post
    PQR is a right angled triangle with PR = 3cm and QR = 4cm. The square STUV is inscribed in triangle PQR. What is the length, in centimeters of the side of the square.

    A) 30/17
    B) 12/7
    C) 5/3
    D) 60/37
    E) 60/39

    Please help!! and tell me how you did it..
    Construction: Draw a perpendicular RO from R to the hypotenuse PQ (such that O lies on PQ).

    Area of the triangle PQR = \frac{1}{2} b h

    \frac{1}{2}*5*RO=\frac{1}{2}*4*3

    RO=\frac{12}{5}

    -----------------------------------------------------------------------------
    In \triangle RQP,

    VS \parallel QP

    By basic proportionality theorem,

    \frac{RV}{RQ}=\frac{RS}{RP}=\frac{SV}{PQ}

    \frac{RV}{4}=\frac{RS}{3}=\frac{SV}{5}=k

    So, RV=4k, RS=3k, SV=5k

    So the side of the square is 5k.

    ------------------------------------------------------------------------------
    In \triangle RQO,

    VU \parallel RO

    By basic proportionality theorem,

    \frac{VU}{RO}=\frac{VQ}{RQ}

    \frac{5k}{\frac{12}{5}}=\frac{VQ}{4}

    \frac{25k}{12}=\frac{VQ}{4}

    VQ=\frac{25k}{3}

    -----------------------------------------------------------------------------

    RV+VQ=4\ cm

    4k+\frac{25k}{3}=4

    12k+25k=12

    k=\frac{12}{37}

    So, the side of the square is 5k=\frac{60}{37}\ cm
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    Last edited by alexmahone; August 5th 2009 at 03:39 AM.
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  3. #3
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    This is my first post in the forum, and
    I'm here to provide an extremely complicated method.

    I'll get you an easier one later.

    Be noticed that the units are all omitted.

    As illurstrated we know that

     PR = 3
     QR = 4

    By applying the Pythagoras' theorem in the right-angled triangle,
    we obtain that

     PQ = \sqrt{PR^2 + QR^2}= \sqrt{25} = 5

    Then we assume the length of the side to be  a .
    In  \triangle UVQ , we can see that

     \tan{Q} = \frac{PR}{RQ} = \frac{UV}{UQ}

    Simplification of the equations above gives that

     UQ = \frac{UV*RQ}{PR} = \frac{4a}{3}

    Similarly, in  \triangle PTS , calculation gives that

     PT = \frac{3a}{4}

    Considering that the segment  SV and the segment  PQ are parallel, we know that,

     \angle{SVR} = \angle{Q}

    Then we may see that

     \sin{\angle{SVR}} = \frac{SR}{SV} = \frac{PR}{PQ} = \sin{\angle{Q}}

    which gives that,

     SR = \frac{3a}{5}

    and similarly

     RV = \frac{4a}{5}

    By doing all the calculations above we see that

     S_{\triangle{PRQ}} = \frac{1}{2}PR*RQ
     S_{\triangle{PRQ}} = \frac{1}{2}*[PT*ST+UQ*VU+SR*RV]+a^2

    By solving the equations we finally obtain that

     a = \frac{60}{37}
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  4. #4
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    Okay here it comes the easier one.

    At first we need to find the length of  PQ .
    My work above gives that

     PQ = 5

    Obviously segments  SV and  PQ are parallel, thus

     \cos{\angle{SVR}} = \cos{\angle{Q}} = 0.8
     \cos{\angle{UVQ}} = \sin{\angle{Q}} = 0.6

    Assume the length of the segment  VQ to be  x , then it naturally satisifies that  RV equals to  4-x .

    Hence, we obtain that

     SV = \frac{RV}{\cos{\angle{SVR}}} = VU = VQ*\cos{\angle{UVQ}}

    which gives that

     x = \frac{48}{37}

    Therefore, the length of the side equals to

     \frac{5x}{4} = \frac{60}{37}
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  5. #5
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    Thanks for all your answers!

    alexmahone you wrote this:

    \frac{1}{2}*5*RO=\frac{1}{2}*4*3

    How do you know RO is 3?
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  6. #6
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    Quote Originally Posted by jgv115 View Post
    Thanks for all your answers!

    alexmahone you wrote this:

    \frac{1}{2}*5*RO=\frac{1}{2}*4*3

    How do you know RO is 3?
    Area of \triangle PQR = \frac{1}{2}bh

    Area = \frac{1}{2}*5*RO=\frac{1}{2}*4*3

    We equate these two values of the area to find RO.

    Thus RO=\frac{12}{5}
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  7. #7
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    Do you mean you make it up? You said that half times 5 times RO (the extra line you added) is the area.

    What I don't understand it where the value of RO came from? You substituted 3 for RO but how did that get there?
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  8. #8
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by jgv115 View Post
    Do you mean you make it up? You said that half times 5 times RO (the extra line you added) is the area.

    What I don't understand it where the value of RO came from? You substituted 3 for RO but how did that get there?
    Area of \triangle PQR=\frac{1}{2}PQ*RO

    Also, Area of \triangle PQR=\frac{1}{2}RQ*RP

    So, \frac{1}{2}PQ*RO=\frac{1}{2}RQ*RP

    Now do you understand?
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  9. #9
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    Similar Triangles

    Hello jgv115
    Quote Originally Posted by jgv115 View Post
    PQR is a right angled triangle with PR = 3cm and QR = 4cm. The square STUV is inscribed in triangle PQR. What is the length, in centimeters of the side of the square.

    A) 30/17
    B) 12/7
    C) 5/3
    D) 60/37
    E) 60/39

    Please help!! and tell me how you did it..
    The solutions you've been given are fine, but look a bit complicated, and most have left quite a bit of the working out. May I suggest a simpler way of setting out the answer?

    Basically, all four of the triangles are 3,4,5 right-angled triangles. This means, for example, that the longest side of any of them is \tfrac53 of the shortest, the middle-sized side is \tfrac45 of the longest, and so on. So, if we call the length of the sides of the square x cm, we can work around the various triangles to get an equation for x, like this:

    In \triangle QUV, UQ = \tfrac43x

    In \triangle RSV, SR = \tfrac35x

    In \triangle PST, PS = 3 - SR = 3 - \tfrac35x

    \Rightarrow PT = \tfrac35PS = \tfrac35(3 - \tfrac35x)

    So if we now add up the three lengths along PQ:

    PT + TU + UQ = \tfrac35(3 - \tfrac35x)+x+ \tfrac43x = PQ = 5

    Multiply both sides of this equation by 75\, (= 3\times 5\times 5):

    45(3 - \tfrac35x) + 75x + 75\times\tfrac43x = 5\times 75

    \Rightarrow 135 - 27x + 75x + 100x = 375

    \Rightarrow 148x = 240

    \Rightarrow x = \frac{60}{37}

    Grandad
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