Hello, realintegerz!

A rectangle is inscribed in a semicircle of radius 5 cm.

One side of the rectangle is on the diameter of the semicircle

and two of the vertices of the rectangle are on the semicircle).

1) Draw a model of this situation Code:

5|
* * *
* | * P
* - - - + - - - o
*| | . |*
| | . W|
* | | . | *
--*-+-------+-------+-*--
-5 : - - - L - - - : 5

The equation of P is: .$\displaystyle (\tfrac{L}{2})^2 + W^2 \:=\:5^2$

The area of the rectangle is: .$\displaystyle A \:=\:LW$

2) Suppose the side of the rectangle which is on the diameter is 2 cm long.

What will be the area of the rectangle? If $\displaystyle L = 2\!:\;\;(\tfrac{2}{2})^2 + W^2 \:=\:25 \quad\Rightarrow\quad W^2 \:=\:24\quad\Rightarrow\quad W \:=\:2\sqrt{6}$

Therefore: .$\displaystyle A \:=\:2\cdot2\sqrt{6} \:=\:4\sqrt{6}$ cm²

3) Create a rule which can be used to find the length of the side of rectangle

which is on the diameter of the semicircle when the area of the rectangle is known. Given $\displaystyle A$, find $\displaystyle L.$

We have: .$\displaystyle (\tfrac{L}{2})^2 + W^2 \:=\:5^2 \quad\Rightarrow\quad W \:=\:\frac{\sqrt{100-L^2}}{2} $

. . Then: .$\displaystyle LW \:=\:A \quad\Rightarrow\quad L\cdot\frac{\sqrt{100-L^2}}{2} \:=\:A$

Solve for $\displaystyle L\!:\;\;L\sqrt{100-L^2} \:=\:2a$

Square both sides: .$\displaystyle L^2(100-L^2) \:=\:4A^2 \quad\Rightarrow\quad L^4 - 100L^2 + 4A^2 \:=\:0$

Quadratic Formula: .$\displaystyle L^2 \;=\;\frac{100 \pm\sqrt{100^2 - 4(4A^2)}}{2} \;=\;50 \pm 2\sqrt{625 - A^2} $

And the positive real root is: .$\displaystyle L \;=\;\sqrt{50 + 2\sqrt{625 - A^2}} $

4) Graph the function from #3 and use it

to find possible lengths which will give an area of 24. I'd let my calculator graph it . . .

5) Why would #4 be hard with algebra alone? Hard? . . . I don't think so!

With $\displaystyle A = 24$, we have: .$\displaystyle L \;=\;\sqrt{50 + 2\sqrt{625 - 24^2}} \;=\;\sqrt{50 + 2\sqrt{49}}$

. . $\displaystyle L \;=\;\sqrt{50 + 2\cdot7} \;=\;\sqrt{50 + 14} \;=\;\sqrt{64}\quad\Rightarrow\quad L \;=\;8 $