Here is a geometry practice problem involving an inscribed circle and rectangle. This type of setup is common on the DAT, and recognizing that the width becomes the diameter of the circle is the key step. Full solution included below.
Problem:
A circle is inscribed within a rectangle so that the two longer sides of the rectangle are tangent to the circle. If the area of the rectangle is 64 and the length of the rectangle is 4 more than three times the width, what is the area of the circle?
Choices:
A) 2π
B) 4π
C) 8π
D) 16π
E) 10π
Answer: B (4π)
Solution:
Let the width be W and the length be L. We know:
- L = 4 + 3W
- The area is W * L = 64
Substitute L:
W * (4 + 3W) = 64
This gives a quadratic equation:
3W² + 4W - 64 = 0
Solve for W and find W = 4 (ignoring the negative root since width must be positive).
Now W = 4, which is the diameter of the circle. So the radius is 2. The area of the circle is π * (2²) = 4π.
