Probe a simple equation

Question

$L$ is the side of a square, and it is 10cm.

In the exercise I have a square shape which is cut with a semicircle in the bottom, and the radius of the semicircle is half the side of the square.

I need steps to prove that

$\displaystyle\frac{\pi L^2}{8} = \frac{\pi r^2}{2}$ (semicircle)

I eventually come to the place where both equations stay equal, 1/8 * Phi * L ² = 1/8 * Phi * L ², but Id like to know which steps I need to make the equation something like 1 = 1, so there's nothing more to do.

Thanks a lot.

It's very simple, because $r = \frac{L}{2}$, you can just plug that in anywhere. — Zetrova, May 02 '14 at 23:22

AsdrubalBeltran · Accepted Answer · 2014-05-02T23:47:02.630

0

Note that, $\displaystyle r=\frac{L}{2}$ then replacement $r$, in $\displaystyle\frac{\pi r^2}{2}$ you obtain $\displaystyle\frac{\pi L^2}{8}$

Know that, the semicircle area is $\displaystyle\frac{\pi r^2}{2}$ how $\displaystyle r=\frac{L}{2}$ then:

$$\frac{\pi r^2}{2}=\frac{\pi (L/2)^2}{2}=\frac{\pi (L)^2}{4\cdot2}=\frac{\pi (L)^2}{8}$$

and is solved.

edited May 02 '14 at 23:47

answered May 02 '14 at 23:18

AsdrubalBeltran

4,212

Is there a way to get keep simplifying and get 1 = 1 or soemthing like that? – JorgeeFG May 02 '14 at 23:31
to demonstrate what they ask, you should start on one side of the equation to obtain the other – AsdrubalBeltran May 02 '14 at 23:38
Could you show me? That is what I cannot do – JorgeeFG May 02 '14 at 23:40
Ok I edit my answer in one moment – AsdrubalBeltran May 02 '14 at 23:42
I came to that solution, should it be finished there or can it go further to the point you delete all variables and stay with a constant like 1 = 1? – JorgeeFG May 02 '14 at 23:51
No, the problem is that you use the equality in the proof, and this is a mistake, for example if we want to prove that 2 * 2 = 4 and change terms then we come to that 1 = 1 or 0=0, to right thing is to start 2 * 2 and reach which equals 4. $ – AsdrubalBeltran May 03 '14 at 00:00

Probe a simple equation

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