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DS Section 16, #17

This topic has 3 member replies
jaheer Junior | Next Rank: 30 Posts Default Avatar
Joined
15 Feb 2007
Posted:
11 messages

DS Section 16, #17

Post Mon Feb 19, 2007 6:38 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    17. The figure above shows the shape of a flower bed. If arc QR is a semicircle and PQRS is a rectangle with QR > RS, what is the perimeter of the flower bed ?
    (1) The perimeter of rectangle PQRS is 28 feet.
    (2) Each diagonal of rectangle PQRS is 10 feet long.

    I've attached the picture. It is a rectangle PQRS with Semi-circle on the top.

    I feel the answer is B, but correct answer mentioned is C. Can someone please explain.

    Thanks in Advance.
    Jaheer

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    Vasudha Senior | Next Rank: 100 Posts Default Avatar
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    Post Sun Mar 11, 2007 7:07 pm
    Hi

    I dont see the attached image ..

    jayhawk2001 Community Manager
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    Post Sun Mar 11, 2007 8:20 pm
    I can't find the image but given that we have the semicircle on top of
    the rectange, lets assume

    PS = QR = l
    RS = QP = w

    We should find l and w to solve this.

    1 - insufficient. All we know is 2(l + w) = 28

    2 - insufficient. All we know is l^2 + w^2 = 100. Cannot solve for l or w.

    Using 1 and 2, we have 2 equations with 2 variables. So, a solution
    should exist and hence is sufficient.

    Sub l = 14 - w

    196 + w^2 - 28w + w^2 = 100
    w^2 - 14w + 48 = 0
    w = 8 or 6

    We are given that l > w and so we know what l is. Perimeter of the
    flower-bed can hence be computed.

    Vasudha Senior | Next Rank: 100 Posts Default Avatar
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    Post Sun Mar 11, 2007 8:50 pm
    Thank you, jayhawk2001.

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