Question

A slice of pizza whose edges form a 42degrees angle with an outer crust edge 4 inches long was found in a gym locker. What was the diameter of the original​ pizza?

Answer 1

Answer:

10.9 inches.

Step-by-step explanation:

We have been given that a slice of pizza whose edges form a 42 degrees angle with an outer crust edge 4 inches long was found in a gym locker. We are asked to find the diameter of the original pizza.

We will use arc length formula to solve our given problem.

$s=r\theta$, where,

s = Perimeter or circumference of circle,

r = Radius,

$\theta$ = Angle in radians.

Let us convert 42 degree to radians.

$42^{\circ}\times\frac{\pi}{180^{\circ}}=\frac{7\pi}{30}$

$4=\frac{7\pi}{30}\cdot r$

$r=\frac{120}{7\pi}$

$r=5.45674090$

We know that diameter is two times radius, so

$\text{Diameter}=5.45674090\times 2$

$\text{Diameter}=10.9134818\approx 10.9$

Therefore, the diameter of the original​ pizza was approximately 10.9 inches.

Answer 2

Answer: Diameter of the pizza is 34.2857142858 inches.

Given that the central angle = θ = 42 degree = 42/180 radian.

The arc length = s = 4 inches.

Let, radius = r.

We have to find the diameter = 2r = ?

Using the formula s = rθ we get:

$4=r\cdot\frac{42}{180}\\r=4\cdot\frac{180}{42}\\r=17.1428571429\\2r=34.2857142858$

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