SOLUTION: A square of side length 10cm is inscribed in a circle. Calculate the diameter of the circle

Question 124499: A square of side length 10cm is inscribed in a circle. Calculate the diameter of the circle
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
It might help you to visualize this problem if you made a rough sketch of it. Draw a circle and
then, on the inside of the circle draw a square that has each of its corners on the circle.
.
Once you do that, draw a diagonal line that connects one corner of the square with the other
corner on the opposite side. This diagonal line will be a diameter of the circle.
.
Notice that this diagonal line also divides the square into two right triangles. You can see
that the two legs of each triangle are sides of the square. Therefore, each leg is 10 cm in length.
You then need to find the hypotenuse (long side) of the triangle and when you do, you will have
the length of the diameter of the circle.
.
You can find the length of the hypotenuse by applying the Pythagorean theorem which says that
the sum of the squares of the two legs of a right triangle is equal to the square of the
hypotenuse. In equation form this is:
.
A^2 + B^2 = C^2
.
where A and B are the two legs and C is the hypotenuse.
.
For this problem, A and B are both 10 cm. So, substituting 10 for A and 10 for B makes the
equation:
.
10^2 + 10^2 = C^2
.
and since 10^2 is equal to 100 you can reduce the equation to:
.
100 + 100 = C^2
.
Add the two numbers on the left side and you have:
.
200 = C^2
.
You can solve for C by taking the square root of both sides to get:
.

.
Just to get this in a conventional form, let's reverse the sides of the equation to get:
.

.
But 200 can be replaced by 100*2 to get:
.

.
and by the rules of square roots this becomes:
.

.
and since the square root of 100 is 10, this further reduces to:
.

.
The units being used are cm, so the answer is that C, the diagonal of the
circle, is:
.
cm
.
Hope this helps you to understand the problem a little better.
.

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