Question

Martin ordered a pizza with a 16-inch diameter. Ricky ordered a pizza with a 20-inch diameter. What is the approximate difference in area of the two pizzas?

Answer 1

Answer:

The surface difference of the two pizzas is 113.14 in².

Step-by-step explanation:

Martin ordered a pizza with a diameter of 16 inches and Ricky ordered a pizza with a diameter of 20 inches.

We need to calculate the approximate difference in the area of the two pizzas.

Ara of pizza ordered by Martin = $\pi r^2$

                                                 = $\pi (\dfrac{16}{2} )^{2} =\pi\cdot 8^{2} =64\pi$

Area of pizza ordered by Ricky = $\pi (\frac{20}{2})^{2} =\pi 10^{2} =100\pi$

Difference in area = $100\pi -64\pi$

                            = $36\pi$

                            = 113.14 inch²

The difference in area of the two pizzas is 113.14 inch².

Answer 2

The approximate difference in the areas of the two pizzas will be 113.04 in².

Given,

Diameter of pizza ordered by Martin = 16 in,

Diameter of pizza ordered by Ricky = 20 in.

To find,

Approximate difference in the areas of the two pizzas.

Solution,

First of all, as we know that a pizza is in a circular shape usually.

So, the area of a pizza can be determined using the following formula.

$A=\pi \frac{d^{2} }{4}$

where d is the diameter of the pizza.

Now, the area of the pizza ordered by Martin (let $A_M$),

$A_M=\pi \times \frac{(16)^{2} }{4}$

Taking $\pi =3.14$ and simplifying,

$A_M=3.14 \times \frac{256}{4}$

$\implies A_M=3.14 \times 64$

$\implies A_M=200.96$ in².

Area of the pizza ordered by Ricky (let $A_R$),

$A_R=\pi \times \frac{(20)^{2} }{4}$

$\implies A_R=3.14 \times \frac{400}{4}$     [again, taking $\pi =3.14$]

$\implies A_R=3.14 \times 100$

$\implies A_R=314$ in².

Now, the difference between the areas of the two pizzas will be $A_R-A_M$.

Thus,

$A_R-A_M=314-200.96$

$\implies A_R-A_M=113.04$ in².

Therefore, the approximate difference in the areas of the two pizzas will be 113.04 in².

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