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Thread: Triangle w/semicircle, calculating area, changing variables in length

  1. #1
    Member realintegerz's Avatar
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    Triangle w/semicircle, calculating area, changing variables in length

    I'm reviewing for finals but this geometry question got me.


    The figure above is an isosceles right triangle with a semicircle attached

    a) Show that the area of the figure is about 5.1 square centimeters when L = 2 cm


    b) Show that L is about 3.9 cm when the area of the figure is 20 cm

    c) Express the distance L as a function of the area of the figure

    This is my work so far, but I didn't get full points on it on the test we had way back. Can anyone point out what I did wrong?

    A) r = sq. root (2^2 + 2^2) = sq. root 8 = 2(sq. root 2)

    Area of triangle = 1/2 (2 sq. root 2)(sq. root 2) = 2
    Area of semicircle = [pi(2 sq. root 2)^2]/2 = 12.5

    Total area = 14.5 square cm

    B) 20 = 1/2(sq. root 2l^2)(sq. root 2l^2/2) + [pi(sq. root 2l^2)2]/2
    20 = 1/2 (2l^2/2) + [pi(2l^2)]/2
    80 = 2l^2 + 2pi(2l^2)
    2l^2 = 10.9
    L = 2.3 cm

    C) A(L) = 2l^2/4 + pi(2l^2)/2
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  2. #2
    No one in Particular VonNemo19's Avatar
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    I assume that the triangle in question is equilateral.

    L=2 implies r=1, therefore

    We take the area of the triangle and add it to the area of the half circle.

    area of triangle equals 1/2(bh) therefore $\displaystyle \frac{1}{2}*2*\sqrt{2^2-1^2}=\sqrt{3}$

    and

    Area of a circle equals $\displaystyle \pi{r^2}$ so $\displaystyle \pi(1^2)=\pi$

    so adding we get $\displaystyle \sqrt{3}+\pi$

    I don't have a calculator, but I;m sure that you got it
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  3. #3
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    But the OP said that the triangle is a right isosceles triangle, so I'm not sure we can assume that it is also equilateral.

    If $\displaystyle L = 2$, then $\displaystyle 2r = 2\sqrt{2}$, meaning that $\displaystyle r = \sqrt{2}$.

    Area of a circle is $\displaystyle A = \pi r^2$, so the area of the semicircle is $\displaystyle \frac{1}{2}\pi r^2 = \frac{1}{2}\pi(\sqrt{2})^2 = \pi$.

    Area of the triangle is $\displaystyle \frac{1}{2}bh = \frac{1}{2}(2)(2) = 2$.

    Add the two areas together to get $\displaystyle \pi + 2 \approx 5.14$.


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    b) Suppose L is unknown. Then by the Pythagorean theorem, the relationship between r and L is
    $\displaystyle \begin{aligned}
    (2r)^2 &= L^2 + L^2 \\
    4r^2 &= 2L^2 \\
    r^2 &= \frac{L^2}{2} \\
    r &= \frac{L}{\sqrt{2}} \\
    r &= \frac{L\sqrt{2}}{2} \\
    \end{aligned}$

    The area of the triangle is $\displaystyle \frac{1}{2}bh$, or $\displaystyle \frac{1}{2}L^2$. The area of the semicircle is $\displaystyle \frac{1}{2}\pi r^2$, so plug in $\displaystyle \frac{L\sqrt{2}}{2}$ for $\displaystyle r$:

    $\displaystyle \begin{aligned}
    \frac{1}{2}\pi r^2 &= \frac{1}{2}\pi \left(\frac{L\sqrt{2}}{2}\right)^2 \\
    &= \frac{1}{2}\pi \left(\frac{L^2}{2}\right) \\
    &= \frac{\pi}{4}L^2
    \end{aligned}$

    The combined area of the semicircle and triangle is

    $\displaystyle \begin{aligned}
    A &= \frac{1}{2}L^2 + \frac{\pi}{4}L^2 \\
    &= \frac{\pi + 2}{4}L^2
    \end{aligned}$

    The area given in part b) is 20 squared cm, so solve:

    $\displaystyle \begin{aligned}
    20 &= \frac{\pi + 2}{4}L^2 \\
    15.559 &\approx L^2 \\
    L &\approx 3.945
    \end{aligned}$


    c) Express the distance L as a function of the area of the figure
    Are you sure this is the question? Because what you wrote is the area A as a function of the length L. If you're looking for the area A as a function of L, I already wrote it above:
    $\displaystyle A = \frac{\pi + 2}{4}L^2$

    But if you truly want the length L as a function of A, solve for L:
    $\displaystyle \begin{aligned}
    A &= \frac{\pi + 2}{4}L^2 \\
    \left(\frac{4}{\pi + 2}\right)A &= L^2 \\
    L &= 2\sqrt{\frac{A}{\pi + 2}}
    \end{aligned}$


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