Everyone in this world will be happy if she/he gets more of an item at the same price. If you go to a pizza joint and you get an offer to choose from two 5-inch pizzas or one 9-inch pizza for the same price, which one will you choose?

Most people will go for two $5$-inch pizzas. But if a person at the counter suggests you go for one $9$-inch pizza, is she/he cheating on you?

To see which one is a better deal, let’s understand the math behind it.

## How Much Does a Pizza Contain?

As you know a pizza has a shape of a disc and a disc can be considered a thin right circular cylinder. The amount of pizza of radius $r$ and thickness $h$ will be the volume of the disc/cylinder of radius $r$ and height $h$.

You might be knowing that Volume = Area $\times$ Height (Thickness)

So, amount you get in an $r$-inch pizza of thickness $h$ will be $ \left( \pi \times r^{2} \right) \times h$.

Now, we’ll use this formula to find how much is present in a $9$-inch pizza and a $5$-inch pizza.

## Two 5-Inch Pizzas – Two 5-Inch Pizzas Or One 9-Inch Pizza

Volume of one $5$-inch pizza = $ \left( \pi \times 5^{2} \right) \times h = \left( 3.14 \times 25 \right) \times \left( 0.5 \right)$ (Considering thickness of a pizza as $0.5$ (half inch))

= $39.25 inch^{3} $

So, two $5$-inch pizzas will have a combined volume of $2 \times 39.25 = 78.5 inch^{3} $.

## One 9-Inch Pizza – Two 5-Inch Pizzas Or One 9-Inch Pizza

Volume of one $9$-inch pizza = $ \left( \pi \times 9^{2} \right) \times h = \left( 3.14 \times 81 \right) \times 0.5$ (Again considering thickness of a pizza as $0.5 inch$ (half inch))

= $127.17 inch^{3}$

## Which One Will You Choose – Two 5-Inch Pizzas Or One 9-Inch Pizza?

As seen from the above calculations –

Two $5$-inch pizzas have a combined volume of $78.5 inch^{3}$.

One $9$-inch pizza has a volume of $127.17 inch^{3}$.

The volume of a $9$-inch pizza is more than the combined volume of two $5$-inch pizzas by $48.67 in^{3} $. $ \left(127.17 – 78.5 \right)$.

Which one will you choose now?

Let’s apply some more math,

Difference in volumes of two cases = $48.67 inch^{3}$

And now finding percentage more by using the formula $ \frac {New – Original}{Original} \times 100$, we get

$ \frac{127.17 – 78.5}{78.5} \times 100 = 62 \%$

**It means one $9$-inch pizza is $62 \%$ more than two $5$-inch pizzas!**

## Let’s Code With Python

Compare the Circumference and Area of Circles

#Accept Data r1 = float(input('Enter radius of first circle ')) r2 = float(input('Enter radius of second circle ')) PI = 22/7 #Value of PI #Compute Circumference C1 = 2 * PI * r1 C2 = 2 * PI * r2 #Compute Area A1 = PI * (r1 ** 2) A2 = PI * (r2 ** 2) #Difference dC = 0 dCP = 0 dA = 0 dAP = 0 if r1 > r2: dC = C1 - C2 dCP = (dC/C1) * 100 dA = A1 - A2 dAP = (dA/A1) * 100 print('Circumference of circle 1 is greater than circumference of circle 2 by', "{:.2f}".format(dC), 'unit') print('Circumference of circle 1 is greater than circumference of circle 2 by', "{:.2f}".format(dCP), '%') print('Area of circle 1 is greater than area of circle 2 by', "{:.2f}".format(dA), 'square unit') print('Area of circle 1 is greater than circumference of circle 2 by', "{:.2f}".format(dAP), '%') elif r1 < r2: dC = C2 - C1 dCP = (dC/C2) * 100 dA = A2 - A1 dAP = (dA/A2) * 100 print('Circumference of circle 2 is greater than circumference of circle 1 by', "{:.2f}".format(dC), 'unit') print('Circumference of circle 2 is greater than circumference of circle 1 by', "{:.2f}".format(dCP), '%') print('Area of circle 2 is greater than area of circle 1 by', "{:.2f}".format(dA), 'square unit') print('Area of circle 2 is greater than circumference of circle 1 by', "{:.2f}".format(dAP), '%') else: dC = 0 dA = 0 print('Circumference of circle 1 is same as circumference of circle 2') print('Area of circle 1 is same as area of circle 2')

## Conclusion

It’s not always what seems to be more is actually more, if we understand the math behind it. The same holds true for successive discounts offered by shop owners to attract customers. (Recommended reading: Understanding Successive Discount).

## Practice Problems

- Which of these has a greater volume – a cube of length $2 cm$ or a sphere of radius $2 cm$?
- A cube of length $6 cm$ is sliced into two equal halves by a plane parallel to one of its faces. Which of the following is true?
- The combined volume of the two solids will be greater than the volume of the original cube
- The combined volume of the two solids will be less than the volume of the original cube
- The combined volume of the two solids will be equal to the volume of the original cube
- The combined total surface area of the two solids will be greater than the total surface area of the original cube
- The combined total surface area of the two solids will be less than the total surface area of the original cube
- The combined total surface area of the two solids will be equal to the total surface area of the original cube

- If the radius of a sphere is increased by $10\%$, then find
- increase in its surface area
- increase in its volume

- A sphere is sliced into two equal halves. Find the increase/decrease percent of the total surface area of the two hemispheres combined with respect to the total surface area of the original sphere.

## Conclusion

It’s not always what seems to be more is actually more, if we understand the math behind it. The same holds true for successive discounts offered by shop owners to attract customers. (Recommended reading: Understanding Successive Discount).