A rectangle is inscribed in a circle with radius 2.5 cm. The perimeter...
The diagonal of the rectangle is diameter of the circle. It is 5 cm.
Let ‘a’ and ‘b’ be the two sides of the rectangle.
a2 + b2 = 25
Now, perimeter = 2(a + b) = 14
a + b = 7
(a + b)2 = 49
ab = 12
Area of Rectangle = ab = 12 cm2 Answer: 12
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A rectangle is inscribed in a circle with radius 2.5 cm. The perimeter...
Problem Analysis:
- We are given that a rectangle is inscribed in a circle with a radius of 2.5 cm.
- The perimeter of the rectangle is 14 cm.
Key Points:
- The diagonal of the rectangle is equal to the diameter of the circle.
- The perimeter of the rectangle is equal to the sum of all four sides.
- The length and width of the rectangle are the two sides adjacent to the diagonal.
Solution:
Let's assume the length of the rectangle is L and the width is W.
Step 1: Determine the diagonal of the rectangle
- The diagonal of the rectangle is equal to the diameter of the circle.
- Therefore, the diagonal = 2 * 2.5 cm = 5 cm.
Step 2: Determine the perimeter of the rectangle
- The perimeter of the rectangle is equal to the sum of all four sides.
- Perimeter = 2L + 2W = 14 cm.
Step 3: Determine the length and width of the rectangle
- Since the length and width are adjacent to the diagonal, we can use the Pythagorean theorem to relate them.
- According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
- In this case, the diagonal is the hypotenuse, and the length and width are the other two sides.
- Therefore, L^2 + W^2 = 5^2.
- Since we have two variables and one equation, we need another equation to solve for L and W.
Step 4: Use the perimeter equation to find the second equation
- Perimeter = 2L + 2W = 14 cm.
- We can rearrange this equation to get L in terms of W: L = (14 - 2W)/2 = 7 - W.
Step 5: Substituting the value of L in the Pythagorean theorem equation
- Substituting the value of L in the Pythagorean theorem equation: (7 - W)^2 + W^2 = 5^2.
- Expanding and simplifying the equation: W^2 - 14W + 49 + W^2 = 25.
- Combining like terms: 2W^2 - 14W + 24 = 0.
- Dividing the equation by 2: W^2 - 7W + 12 = 0.
- Factoring the quadratic equation: (W - 3)(W - 4) = 0.
- Solving for W: W = 3 or W = 4.
Step 6: Determine the length and width of the rectangle
- If W = 3, then L = 7 - 3 = 4.
- If W = 4, then L = 7 - 4 = 3.
Step 7: Calculate the area of the rectangle
- The area of the rectangle = length * width.
- If W = 3, then the area = 4 * 3 = 12 cm^2.
- If W = 4, then the area = 3
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