maths help?

Question

i cant solve this question ..

If the diameter of the pizza was 50cm, and the half-pizza was a perfect semicircle, what is the area of the smallest square box, in square centimetres, that could contain the half pizza?

disregared the thickness of the box and round to the nearest whole number.

Answer 1

Put the pizza in the box so that the diagonal of the box lines up with the diagonal of the pizza.

Now the distance from the center of the pizza box to the edge of the pizza box needs to be equal to the radius of the pizza, which is 25 cm.

So the area of the box is 50 cm * 50 cm = 2500 cm^2

Cutting the pizza in half uses the same size square box as leaving it whole. If you can cut the half-pizza in half again, you can turn the center toward the corner and the other center toward the other corner. Then you can fit the half pizza into a box with a diagonal of 50 cm.

The sides would be 2s^2 = 50^2
s = 35.355 cm
s^2 = 1250 cm^2

If you cut the pizza into smaller and smaller slices, you can get down to a box that has an area of 1/2 pi* r^2 =
1/2 pi * 25^2 = 981.7 cm^2 rounds to 982 cm^2

Answer 2

If you have to keep it in a semicircle, then the box would have to be 50cm wide to hold it. If the box has to be square, then it would be 50cm x 50cm = 2500 cm^2.

If you're allowed to rearrange it, it depends on how you do the rearranging. The area of the pizza will be (1/2) pi (25 cm)^2 = 982 cm^2, so that would be the absolute smallest box possible. To do that, you'd have to cut the thing up like a puzzle to fit it in.

Given the way the question is asked, that's about all I can tell you. The box would have to be between 982 cm^2 and 2500 cm^2, depending on how or if you're allowed to rearrange the pizza.

Answer 3

diameter times radius equals 25x50=1250cm^2

unless the box can be a semicircle then you need the pie stuff and I don't do that