Question

Problem 5. A skating rink in the shape shown has an area of

Answer 1

Answer:

$P=\dfrac{\pi r^2+2800}{r} ft$

Step-by-step explanation:

Let the length of the rectangular part =l

The width will be equal to the diameter of the semicircles.

Area of the Skating Rink= $2(\frac{\pi r^2}{2})+(lX2r)$

Therefore:

$\pi r^2+2lr=2800\\2lr=2800-\pi r^2\\ Divide both sides by 2r\\l=\dfrac{2800-\pi r^2}{2r}$

Perimeter of the Shape =Perimeter of two Semicircles + 2l

$=2\pi r+2\left(\dfrac{2800-\pi r^2}{2r}\right)\\=2\pi r+\dfrac{2800-\pi r^2}{r}\\=\dfrac{2\pi r^2+2800-\pi r^2}{r}\\=\dfrac{\pi r^2+2800}{r}$

The perimeter of the rink is given as:

$P=\dfrac{\pi r^2+2800}{r} ft$