1. ## Rectangle inside semicircle

I have no clue what to do here......

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

2) Suppose the side of the rectangle which is on the diameter is 2 cm long. What will be the area of the rectangle?

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

4) Graph the function from #3 and use it to find possbiel lengths which will give an area of 24

5) Why would #4 be hard with algebra alone?

1) See attached diagram. The radius of semicircle = r = 5 cm
Let the sides of rectangle be x and y. The centre O of semi-circle divide side x into two equal parts.
AO = OD = x/2
In rectangle ABCD, AD = BC = x cm, AB = CD = y cm.
OB = OC = radius = 5 cm.

2) Side of rectangle, x = 2 cm.
In triangle ABO,
AO = x/2 = 2/2 = 1 cm

Apply pythagorean theorem in triangle ABO,

$\displaystyle AB^2+AO^2=OB^2$

$\displaystyle y^2+ \left( \frac{x}{2}\right)^2=r^2$

$\displaystyle y^2+ (1)^2=5^2$

$\displaystyle y^2=25- 1=24$

$\displaystyle y=\sqrt{24}=4\sqrt{6}$

Area of rectangle ABCD $\displaystyle = x.y = (1)4\sqrt{6} = 4\sqrt{6} \;cm^2$

3)Apply pythagorean theorem in triangle ABO,

$\displaystyle AB^2+AO^2=OB^2$

$\displaystyle y^2+ \left( \frac{x}{2}\right)^2=5^2$

$\displaystyle y^2 =25-\frac{x^2}{4}$

$\displaystyle \boxed {y =\frac{\sqrt{100-x^2}}{2}}$

Area of rectangle $\displaystyle =x.y =x.\frac{\sqrt{100-x^2}}{2}=\frac{\sqrt{100x^2-x^4}}{2}$

$\displaystyle Area = A = \frac{\sqrt{100x^2-x^4}}{2}$

squaring both sides,

$\displaystyle (A)^2 = \frac{100x^2-x^4}{4}$

$\displaystyle \boxed{x^4-100x^2 + 4A^2=0}$

4) $\displaystyle x^4-100x^2 + 4A^2=0$

$\displaystyle x^4-100x^2 + 4(24)^2=0$

$\displaystyle x^4-100x^2 + 2304 = 0$

$\displaystyle (x^2-64)(x^2 -36) = 0$

x = -6, 6, -8, 8

x = 6 cm and x = 8 cm. because, side of rectangle > 0.

so, side length = 6 cm or 8 cm for area $\displaystyle 24 \;cm^2$

3. ## Graph

4) Please see attached graph for

$\displaystyle \boxed{x^4-100x^2+4A^2=0, \;x>0, \; A>0.}$

for A = 24, we get x = 6, 8

side of rectangle = 6 cm and 8 cm

4. 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: .$\displaystyle L \;=\;\sqrt{50 + 2\sqrt{625 - 24^2}} \;=\;\sqrt{50 + 2\sqrt{49}}$

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

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# abcd is a rectangle inscribed in a circle of radius 5 cm

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