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

**realintegerz** · June 7th 2009, 06:31 PM

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

**VonNemo19** · June 7th 2009, 07:45 PM

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

**yeongil** · June 7th 2009, 09:32 PM

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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**yeongil** · June 7th 2009, 09:48 PM

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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