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Math Help - Find radius of circle

  1. #1
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    Find radius of circle

    The only information provided is the lengths of the sides of the triangle

    Question: Find the radius of the circle. Ans:2cm

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  2. #2
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    Quote Originally Posted by Punch View Post
    The only information provided is the lengths of the sides of the triangle

    Question: Find the radius of the circle. Ans:2cm

    Draw lines connecting the point where the circle touches the triangle and the center. Also, draw a line from C to the center of circle.

    Use the fact that both tangents from a point to the same circle are equal.
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  3. #3
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    hi, i tried that and the answer is 2.6... which is wrong!!!!

    we probably couldnt assume that the lengths of the triangle are tangents as it is not stated in the question
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    Quote Originally Posted by Punch View Post
    hi, i tried that and the answer is 2.6... which is wrong!!!!

    we probably couldnt assume that the lengths of the triangle are tangents as it is not stated in the question
    Can you show us your computations?
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  5. #5
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    Angle ACB=sin^-1\frac{5}{13}<br /> <br />
=22.6198

    Let length of radius be  x and point of centre of circle be O

    Angle OCA=\frac{22.6198}{2}
    =11.3099 (external point of tangent of circle)

    tan11.3099=\frac{x}{13}

    x=13tan11.3099
    =2.6cm
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  6. #6
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    Quote Originally Posted by Punch View Post


    Angle ACB=sin^-1\frac{5}{13}<br /> <br />
=22.6198

    Let length of radius be  x and point of centre of circle be O

    Angle OCA=\frac{22.6198}{2}
    =11.3099 (external point of tangent of circle)

    tan11.3099=\frac{x}{13} ratio is incorrect (not 13)

    note that the area of triangle ABC = 30

    sketch segments from the circle's center to each vertex. three triangles are formed.

    \frac{1}{2}(5 \cdot r) + \frac{1}{2}(12 \cdot r) + \frac{1}{2}(13 \cdot r) = 30

    solve for r
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  7. #7
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    Quote Originally Posted by Punch View Post


    Angle ACB=sin^-1\frac{5}{13}<br /> <br />
=22.6198

    Let length of radius be  x and point of centre of circle be O

    Angle OCA=\frac{22.6198}{2}
    =11.3099 (external point of tangent of circle)

    tan11.3099=\frac{x}{13}

    Check this again, it's not over 13.

    x=13tan11.3099
    =2.6cm
    The circle does touch the triangle. Else, it can't be solved.

    Lets call the point of intersection (for the one above ) be P and the one below, Q, and the one by the side, R.

    AP=AR=x and RB=BQ=y

    CP=13-x and CQ=12-y

    x+y=5
    13-x=12-y

    Solving the simultaneous equation yields x=3, y=2.

    angle ACB=22.62

    angle COP=11.3

    tan 11.3 = radius / 10

    radius =2
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  8. #8
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    thanks i overlooked that
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  9. #9
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    Quote Originally Posted by skeeter View Post
    note that the area of triangle ABC = 30

    sketch segments from the circle's center to each vertex. three triangles are formed.

    \frac{1}{2}(5 \cdot r) + \frac{1}{2}(12 \cdot r) + \frac{1}{2}(13 \cdot r) = 30

    solve for r
    i didnt quite understand the points u said to draw

    the part where u said to each vertex
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