Consider only the rectangle for a moment. Diagonal BD cuts the rectangle into two right triangles, and the length of this diagonal is given:
Now look at right triangle ABD. The line segment BD functions not only as the diagonal of rectangle ABCD but also as the hypotenuse of right triangle ABD. So now find the third side of triangle ABD, either using the Pythagorean theorem or recognizing a Pythagorean triple (6–8–10):
Now consider the circle within this 6 by 8 rectangle:
Finally, find the radius and compute the area:
d = 6 = 2r 3 = r |
Area = πr2 Area = π32 Area = 9π |