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Harlamova29_29 [7]
3 years ago
13

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

Mathematics
1 answer:
nexus9112 [7]3 years ago
4 0

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

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