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Math Help - [SOLVED] SAT geometry

  1. #1
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    [SOLVED] SAT geometry

    In the figure below, arc SBT is one quarter of a circle with center R and radius 6. If the length plus the width of rectangle ABCR is 8, then the perimeter of the darkened region is

    a) 8+3pi

    b) 10+3pi

    c) 14+3pi

    d) 1+6pi

    e) 12+6pi

    Can you please show the steps? Thanks in advance

    hey i forgot a major info. The graph stated that SR is 6, sorry i didn't include it on the graph!!
    Attached Thumbnails Attached Thumbnails [SOLVED] SAT geometry-738.jpg  
    Last edited by fabxx; September 29th 2008 at 12:43 AM. Reason: mistake ^.^
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    Quote Originally Posted by fabxx View Post
    In the figure below, arc SBT is one quarter of a circle with center R and radius 6. If the length plus the width of rectangle ABCR is 8, then the perimeter of the darkened region is

    a) 8+3pi

    b) 10+3pi

    c) 14+3pi

    d) 1+6pi

    e) 12+6pi

    Can you please show the steps? Thanks in advance
    i will start you off.

    Let RC be w, that is, the width of the rectangle.
    Let AR be l, that is, the length of the rectangle.
    Let P be the perimeter of the darkened region.

    Note that the diagonal AC = 6

    we are given l + w = 8 ............(1)

    by Pythagoras' theorem, l^2 + w^2 = 6^2 .............(2)

    you have two simultaneous equations with two unknowns, you can solve for l and w. the rest should be straight forward

    P = AS + AC + CT + \text{ arc}SBT

    now, AS + l = 6 \implies \boxed{AS = 6 - l}

    \boxed{AC = 6}

    CT + w = 6 \implies \boxed{CT = 6 - w}

    \boxed{\text{arc}SBT = \frac {90}{360} \cdot 2 \pi r}, and you know r.

    so you can find all the components and hence the perimeter
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    Hey i forgot to include a major info in the graph. The radius is 6 is meant that SR is 6. I already corrected it in my previous thread. (:
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fabxx View Post
    Hey i forgot to include a major info in the graph. The radius is 6 is meant that SR is 6. I already corrected it in my previous thread. (:
    as you can see, we already figured that out. you told us that R was the center and the circle had a radius of 6 (RT = 6 also ). nothing changes in what i did
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  5. #5
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    Fabxx,

    This is SATs. Pythagoras is too long for SAT. You are supposed to get this answer in not more than 30 seconds, without a calculator. The worst thing a candidate make in a SAT exam is take it with a Texas Instruments. Really, this is not a joke.

    To solve this question, follow me carefully:
    It is given that AR + RC =8 ...(1)
    SR=RB=CA=6 ...(2)

    Perimeter = SA + arc ST + TC + CA (From above, we already know CA=6)

    SA = 6-AR
    TC= 6 - RC

    Add the two equations above:
    SA + TC = 12 - (AR+RC) = 12 - 8 =4

    Therefore,

    Perimeter = arc ST + 4 + 6 = \frac{1}{4}*2*\pi*6 + 4 +6 = 10+3 \pi


    As you can see, you save time by not solving simultaneous equations and not evaluating pythagoras expression.

    Hope that helps.
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    Quote Originally Posted by shailen.sobhee View Post
    Fabxx,


    Perimeter = arc ST + 4 + 6 = \frac{1}{4}*2*\pi*6 + 4 +6 = 10+3 \pi
    how did you get that arc ST is \frac{1}{4}*2*\pi*6 ?

    Thanks in advance (:
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fabxx View Post
    how did you get that arc ST is \frac{1}{4}*2*\pi*6 ?

    Thanks in advance (:
    2 \pi * 6 gives the circumference. you have 1/4 of it (it's a quarter circle)
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    Thanks to shailen.sobhee and Jhevon!
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  9. #9
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    Quote Originally Posted by shailen.sobhee View Post
    Fabxx,

    SR=RB=CA=6 ...(2)
    How did you get that CA=6? CA isn't the radius, and from the graph, you can easily see that CA≠SR

    Thanks in advance
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fabxx View Post
    How did you get that CA=6? CA isn't the radius, and from the graph, you can easily see that CA≠SR

    Thanks in advance
    you are correct. but RB is the radius, and CA = RB since they are both diagonals of the rectangle (sobhee mentioned this)
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