Area of semicircle + isoceles triangle

**realintegerz** · October 4th 2008, 11:46 AM

I'm trying to find the formula to get the area as a function of that length....

So far I got

Area =

1/2[sq. root(2L^2)][sq. root(L^2 - [(sq. root((2L^2)/2)]^2)] + [pi([(sq. root((2L^2)/2)^2]]/2

The question asks...find the length of l for area 30 cm^2

**icemanfan** · October 4th 2008, 11:52 AM

The radius r of the semicircle is $r = \frac{\sqrt{2}}{2}L$ .

Hence, the area of the semicircle is $\frac{1}{2}\pi{r^2} = \frac{1}{2}\pi \cdot \left(\frac{1}{2}L^2\right) = \frac{1}{4}\pi{L^2}$ .

The area of the triangle is $\frac{1}{2}L^2$ .

So the total area A is $A = \left(\frac{1}{2} + \frac{\pi}{4}\right)L^2$ .

Solve A = 30 for L.

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