What is the area of the smallest square a semi-circle which the flat side is 50cm fit into?

Question

If you have a semi-circle, which the flat side is 50cm, what is the area of the smallest possible square which the semi-circle will fit into? I got 1250cm, but feel as this is wrong. Any help?

Answer 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.

Answer 2

It's hard to put this into words but I'll try to steer you towards the solution. This is a tougher question than it may seem. The semicircle fits into a square. We want the square to be as small as possible so it should "touch" the semicircle at some points. The initial instinct may be to have the base of the semicircle rest on the edge of the square, but we can do better. If you "tilt" the base of the semicircle so that the semicircle is sort of slanted with the endpoints of its base touching 2 adjacent sides of the square, the area will be smaller. The semicircle should be tangent to the square at one point on the square's other two sides. It turns out that if the base of the semicircle forms 45 degree angles with the square's sides, then the area of the square is minimized.

After adding some lines (radii and perpendiculars) and using 45-45-90 relationships, I get that L = 25(1 + 1/rt(2)) where L is the side length of the square. Then, A = L^2 (I'm too lazy to expand that right now)

Note: I didn't actually prove that the square I described above has minimal area. That's a more difficult question which I'll leave for you to ponder. There's an AIME (American Invitational Math Exam) question that's similar to this except it asks for the maximum area of a semicircle fitted into a given square (same idea though i think). I think it may have been the 2005 AIME give or take a couple years.

*Addition*

Check out Puzzling's diagram below.

Answer 3

The circle will touch the square at the middle point of each of the square's sides. So the distance from one of these points to the point on the opposite side is equal to the diameter of the circle and is the length of the sides of the square. Here radius = 3.5 hence diameter = 7, hence the area of the square is 7 x 7 = 49 . Note that you haven't given any units for the length of the radius so you have to give the area simple as 49. Strictly speaking it should be 49 square units of length where the unit of length is the same as that of the radius. That's quite a mouth full. Your teacher could give you extra marks for being mathematically correct or simply think you are a smarty pants.

Answer 4

Well the entire flat side of the semi-circle must be inside, so the shortest that side can be is 50cm. Also, if you have a semi-circle, you essentially have one diameter and one radius, so the diameter (50) will be the longest part. It's a square, so the other sides must also be 50cm. Area will be length x width: 50 x 50 = 2,500 cm^2.

Answer 5

25*25*pi= area of whole pizza
(25*25*pi)/2= area of semi-circle
sqrt((25*25*pi)/2)= side of pizza box(assuming you make a square pizza)

=sqrt((625*3.14)/2)
=sqrt(1962.5/2)
=sqrt(981.25)
=31.3249...

The side of the square that holds the square shaped pizza with the equivalent of a semi-circle with a radius of 25 cm, than the box side would measure 31.325 cm (rounded).

Answer 6

the square will have area 2500 sq. cm but smallest rectangle may have area 1250 sq.cm

Cheers!