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Thread: geometry problem

  1. #1
    Apr 2008

    geometry problem

    Here are some more questions for you math enthuaists.

    I'll explain to you where I ca

    1. An equilateral triangle has a perimeter of 24 cm. Find the area of the circle which is inscribed in the triangle.

    (16pie)/3 cm^2, but how and why?

    My work: I got 3 more equalateral triangles (side lengths =4).

    Got a triangle in the circle. Lengths = 4.

    Right triangle in circle = sqrt(12), 4, 2.

    Now... I'm stuck.

    2. Given a circle inscribed in an isosceles trapezoid bases 8 cm and 18 cm.

    a) The diamter of the circle.
    2. The area of the shaded region.

    I don't knwo how to approach this problem. Or even start it. Class never learned this, but teacher said we should know this by now.
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  2. #2
    Super Member Aryth's Avatar
    Feb 2007
    1. All you need for this one is the length of each side and the semiperimeter (Half of the perimeter). The formula for the semiperimeter is:

    $\displaystyle s = \frac{a + b + c}{2}$

    Where a, b, and c are the sides of the triangle, and since they are equilateral triangles, they all have the same side length, which I will call $\displaystyle l$. Our new formula is:

    $\displaystyle s = \frac{3l}{2}$

    We know that l has to be 8, because it's a three-sided shape with equal side lengths, so our semiperimeter is:

    $\displaystyle s = \frac{24}{2} = 12$

    Now, we need to know what's called the inradius, which we use the following equation to get:

    $\displaystyle r = \sqrt{\frac{(s - a)(s - b)(s - c)}{s}}$

    Now, we know that s is 12 and a b and c are all 8, so we just plug and chug:

    $\displaystyle r = \sqrt{\frac{(12 - 8)(12 - 8)(12 - 8)}{12}}$

    $\displaystyle r = \sqrt{\frac{64}{12}}$

    $\displaystyle r = \sqrt{\frac{16}{3}}$

    $\displaystyle r = \frac{4}{\sqrt{3}}$

    Now, the area of a circle is:

    $\displaystyle A = \pi r^2$

    We know the radius, all we have to do is plug and solve:

    $\displaystyle A = \pi \left(\frac{4}{\sqrt{3}}\right)^2$

    $\displaystyle A = \pi \left(\frac{16}{3}\right)$

    $\displaystyle A = \frac{16\pi}{3} \ cm^2$

    And there's that one.

    2. Notice that the median of the trapezoid fulfills the diameter of circle completely. and the formula for that is:

    $\displaystyle median = \frac{b_1 + b_2}{2}$

    Simple enough:

    $\displaystyle median = \frac{26}{2} = 13 \ cm$

    This is the diameter of the circle, so now we also have the height of the trapezoid and can get the area of the circle. First, we'll get the area of the trapezoid:

    $\displaystyle A = \frac{h(b_1 + b_2)}{2}$

    $\displaystyle A = \frac{13(26)}{2} = 169 \ cm^2$

    Now, we need the area of the Circle:

    $\displaystyle A = \pi r^2$

    $\displaystyle A = \pi (42.25)$

    $\displaystyle A = \pi \left(\frac{169}{4}\right)$

    $\displaystyle A = \frac{169\pi}{4} \ cm^2$

    Now, when you draw the median, you cut the trapezoid into two separate trapezoids, we are interested in the bottom half, where we know the following:

    $\displaystyle b_1 = 13 cm$

    $\displaystyle b_2 = 18 cm$

    $\displaystyle \text{Area of Half-Circle} = \frac{\frac{169\pi}{4}}{2} = \frac{169\pi}{8} \ cm^2$

    $\displaystyle \text{Height of New Trapezoid} = 6.5 \ cm$

    $\displaystyle \text{Area of New Trapezoid} = \frac{6.5(31)}{2} = \frac{403}{4} \ cm^2$

    Now we need to subtract the area of the half-circle from the area of the new trapezoid, and we will be left with the areas of the two identical corners:

    $\displaystyle A = \frac{403}{4} - \frac{169\pi}{8} = 34.384 \ cm^2$

    Now, we divide this new area by 2 to get the area of a single corner:

    $\displaystyle A_{corner} = \frac{34.384}{2} = 17.192 \ cm^2$

    There you go.
    Last edited by Aryth; Jun 2nd 2008 at 08:16 PM.
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