a small man realized he couldnt finish his pizza in fact he could only eat exactly half of it.
If the diameter of the pizza was 50cm, and it was a perfect semicircle, what is the area of the smallest square box, in square centimetres, that could contain the half pizza? Please disregard the thickness of the box, round to the nearest whole number, and please only submit a number for your answer.
2007-10-31 12:22:01 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
thank you everybody...math is not my strongpoint.....and i just copied the question...im glad to know how to do it so i dont have to ask in the future
2007-10-31 12:39:56 · update #1
Imagine starting by putting the diameter of the semicircle directly in the corner of a square but at a 45 degree angle. That makes a 45-45-90 triangle with the hypotenuse being 50 cm.
Now take the midpoint of this and draw your semicircle. The distance from this point up will be the radius (25 cm). And down will be the side of a triangle with hypotenuse 25. This ends up being 25/√2.
So one dimension would be 25 + 25/√2, and so would the other. Multiplying this together we get:
(25 + 25/√2)²
= 25² + 50*25/√2 + (25/√2)²
= 625 + 1250/√2 + 625/2
≈ 625 + 883.883 + 312.5
≈ 1821 sq. cm
I drew the following diagram to help explain things.
2007-11-01 20:59:56 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
The box would have to be a square (since this is a stated requirement) of 50cm on a side.
So area = 2500 cm^2
But you could fit it in a rectangular box of dimensions 50 cm by 25 cm for an area of only 1250 cm^2. But this would violate your requirement that the box be square.
2007-10-31 12:32:51 · answer #2 · answered by ironduke8159 7 · 0⤊ 1⤋
exactly 1250 square centimeters.
why can't I show how to get it?
2007-10-31 12:31:24 · answer #3 · answered by Marley K 7 · 0⤊ 1⤋