Arectangle is inscribed in a semicircle of radius 8 cm. what is the maximum area of the rectangle?

A = 64

Step-by-step explanation:

As we can see from the picture, the diagonal for the triangle is length r.    Label the height x. We can use Pythagorean theorem to calculate the length of the base of the triangle.

r^2 = x^2 +b^2

r^2 -x^2 = b^2

Taking the square root of each side

sqrt(r^2 -x^2)

The width of the rectangle is twice the square root of (r^2-x^2)

and the height is x, so we can find the area

A = x* 2(sqrt(r^2 -x^2))

Substitute r=8  into the equation

A = 2x  sqrt (64 - x^2)

Square both sides

A^2 = 4x^2 (64-x^2)

Distribute

A^2 = 256x^2 - 4x^4

Subtract A^2 from each side

0 = -4x^4 +256x^2 -A^2

Replace x^2 with m, this allows us to find the maximum value of m by find the axis of symmetry

0 = -4m^2 +256m^2 -A^2

The axis of symmetry for m is -b/2a

so the axis of symmetry is -256/(2 *-4)

The axis of symmetry for m is 32

Now we need to substitute back in for x

x^2 = m

x^2 =32

Taking the square root of each side

x = sqrt (32)

x = 4sqrt(2)

The axis of symmetry is 4sqrt(2) for x

The maximum is at a 4 sqrt(2)

Substituting this back into the equation for A

A = 2(4 sqrt(2)  sqrt (64 - (4sqrt(2))^2)

A = 8sqrt(2) * sqrt(64-32)

A = 8 sqrt(2) * sqrt(32)

A = 8sqrt(2) * 4sqrt(2)

A = 32*2

A = 64

See attached picture:

b is the correct answer