# Thread: Rectangle Inscribed inside a Semicircle (w/ picture)

1. ## Re: Rectangle Inscribed inside a Semicircle (w/ picture)

Originally Posted by Binary
What i can tell from what I see in my calculator. It looks like the Area = 25cm^2, But as you stated there are 4 congruent sides, so Im guessing the area = 100cm^2?
no ... I said the rectangle's area equals 4 congruent triangle areas. The area of one triangle is $\frac{1}{2} \cdot 5\cos{\theta} \cdot 5\sin{\theta}$. Multiplying this expression by four gives the rectangle area, $A = 25\sin(2\theta)$.

Note that the maximum area of the rectangle is 25 because the largest value that $\sin(2\theta)$ can be is 1.

$A_{max} = 25 \cdot [the \, largest \, possible \, value \, of \, \sin(2\theta)] = 25(1) = 25$

2. ## Re: Rectangle Inscribed inside a Semicircle (w/ picture)

So you have to look at what Max Area sin(2theta) can be? So once you have that, you substitue 1 for sin(2theta) and that equals 25cm^2, which gives you the max area of the whole rectangle?

3. ## Re: Rectangle Inscribed inside a Semicircle (w/ picture)

I see to find the Dimensions of the rectangle, I know that 2(theta)= 90º or pi/2, (theta)=45º or pi/4. How would I go about solving this part?

-Thank you very much skeeter for all your help by the way. Your the best <3

4. ## Re: Rectangle Inscribed inside a Semicircle (w/ picture)

Originally Posted by Binary
I see to find the Dimensions of the rectangle, I know that 2(theta)= 90º or pi/2, (theta)=45º or pi/4. How would I go about solving this part?

-Thank you very much skeeter for all your help by the way. Your the best <3
you just solved it ...

$\sin(2\theta) = 1$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

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