A circle of radius greater than 9 cm is inscribed in the square ABCD. A point P on the circle is 8 cm from side AB of the square, and 9 cm from side AD. What is the radius of the circle?
D-----------C
| |
| | <---circle inscribed in it
| |
| |
A-----------B
sorry the square got messed up
that's ok -- we know what a square looks like.
Looks like we have
(r-8)^2 + (r-9)^2 = r^2
To find the radius of the circle inscribed in the square, we can use the Pythagorean theorem and the concept of tangent lines to a circle.
Let's start by drawing the square ABCD with the inscribed circle and point P.
D-----------C
| |
| | <---circle
| |
| |
A----P-----B
Since the circle is inscribed in the square, the center of the circle, O, will be the midpoint of diagonal AC.
Next, let's extend line segment OP, passing through point P, until it intersects side AB at point Q.
D-----------C
| |
| ---------Q
| /
| /
A----P---B
Since OP is a radius of the circle and PQ is tangent to the circle at point P, OP will be perpendicular to PQ. This means that triangle OPQ is a right triangle.
We are given that point P is 8 cm from side AB and 9 cm from side AD. Therefore, PQ is the difference between these lengths: PQ = AD - AP = 9 cm - 8 cm = 1 cm.
Now, let's consider triangle OPQ. We have the lengths of two sides, PQ = 1 cm and OQ = radius of the circle (which we want to find). We can use the Pythagorean theorem to find the third side OP.
OP^2 = OQ^2 + PQ^2
Since OP is the length of the diagonal AC, which is equal to the side length of the square, we can substitute AC for OP:
AC^2 = OQ^2 + PQ^2
The side length of the square is equal to the diameter of the circle, so we can substitute 2r for AC:
(2r)^2 = OQ^2 + 1^2
Simplifying, we get:
4r^2 = OQ^2 + 1
Now, we need to find the value of OQ, which is the radius of the circle. To do this, we need to find the value of r that satisfies the given conditions.
Since the circle is inscribed in the square, the radius of the circle will be half the length of the side of the square. Let's call the side length of the square x. Therefore, the radius of the circle will be r = x/2.
Substituting this into our equation, we get:
4(x/2)^2 = OQ^2 + 1
x^2 = 4OQ^2 + 4
But we also know that the side length of the square is greater than 9 cm. Therefore, x > 9 cm.
Substituting x > 9 cm into our equation, we get:
x^2 > 9^2 = 81
4OQ^2 + 4 > 81
4OQ^2 > 77
OQ^2 > 19.25
Since OQ is the radius of the circle, it must be positive, so we take the square root of both sides:
OQ > √19.25
Calculating the square root of 19.25, we find:
OQ > 4.38
Therefore, the radius of the circle is greater than 4.38 cm.
In conclusion, the radius of the circle is greater than 4.38 cm.