# Triangle w/semicircle, calculating area, changing variables in length

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• Jun 7th 2009, 06:31 PM
realintegerz
Triangle w/semicircle, calculating area, changing variables in length
I'm reviewing for finals but this geometry question got me.

http://i40.tinypic.com/jzbj2p.jpg
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
• Jun 7th 2009, 07:45 PM
VonNemo19
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 $\frac{1}{2}*2*\sqrt{2^2-1^2}=\sqrt{3}$

and

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

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

I don't have a calculator, but I;m sure that you got it(Wink)
• Jun 7th 2009, 09:32 PM
yeongil
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 $L = 2$, then $2r = 2\sqrt{2}$, meaning that $r = \sqrt{2}$.

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

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

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

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• Jun 7th 2009, 09:48 PM
yeongil
b) Suppose L is unknown. Then by the Pythagorean theorem, the relationship between r and L is
\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 $\frac{1}{2}bh$, or $\frac{1}{2}L^2$. The area of the semicircle is $\frac{1}{2}\pi r^2$, so plug in $\frac{L\sqrt{2}}{2}$ for $r$:

\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

\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:

\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:
$A = \frac{\pi + 2}{4}L^2$

But if you truly want the length L as a function of A, solve for L:
\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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