Question

Rectangle LMNO is inscribed in a circle with center P. If the area of the rectangle is 8 times its width, and the distance from P to side LM is 3, what is the circle's approximate circumference?

• A. 5
• B. 10
• C. 30
• D. 45
• E. 75

Hint:

A key relationship to remember is that the diagonal of an inscribed rectangle is always a diameter of the circumscribing circle. Visualizing or sketching the problem can help clarify the geometric relationships.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem combines the properties of a rectangle and a circle. The key facts are that the diagonal of a rectangle inscribed in a circle is the diameter of the circle, and the center of the circle is also the center of the rectangle.
Step 2: Key Formula or Approach:
- Area of a rectangle: \( A = \text{Length} \times \text{Width} \)
- Pythagorean theorem for the rectangle's diagonal (\(d\)): \( d^2 = L^2 + W^2 \)
- Circumference of a circle: \( C = \pi \times \text{diameter} \)
Step 3: Detailed Explanation:
1. Find the dimensions of the rectangle.
Let the length of the rectangle be \(L\) and the width be \(W\).
We are given that the area is 8 times its width:
\[ L \times W = 8 \times W \] Dividing both sides by \(W\) (assuming \(W \neq 0\)), we get:
\[ L = 8 \] Now, let's assume side LM is the length, so its measure is 8. The center of the inscribed rectangle, P, is also the center of the circle. The distance from the center P to a side (like LM) is half the measure of the perpendicular side (the width).
We are given that the distance from P to side LM is 3. Therefore:
\[ \frac{W}{2} = 3 \] \[ W = 6 \] So, the rectangle has dimensions Length = 8 and Width = 6.
2. Find the diameter of the circle.
The diagonal of the inscribed rectangle is the diameter of the circle. We can find the length of the diagonal using the Pythagorean theorem with the rectangle's length and width.
Let \(D_{circle}\) be the diameter of the circle, which is equal to the diagonal of the rectangle.
\[ (D_{circle})^2 = L^2 + W^2 \] \[ (D_{circle})^2 = 8^2 + 6^2 \] \[ (D_{circle})^2 = 64 + 36 = 100 \] \[ D_{circle} = \sqrt{100} = 10 \] 3. Calculate the circle's circumference.
The formula for the circumference is \( C = \pi \times D_{circle} \).
\[ C = \pi \times 10 = 10\pi \] The question asks for the approximate circumference. We use the approximation \( \pi \approx 3.14 \).
\[ C \approx 10 \times 3.14 = 31.4 \] Looking at the options, the value 31.4 is closest to 30.
Step 4: Final Answer:
The circle's approximate circumference is 30.