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A rectangle is inscribed in a circle with radius 2.5 cm. The perimeter of the rectangle is 14 cm. Find the area of the rectangle (in cm2).
    Correct answer is '12'. Can you explain this answer?
    Verified Answer
    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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    A rectangle is inscribed in a circle with radius 2.5 cm. The perimeter of the rectangle is 14 cm. Find the area of the rectangle (in cm2).Correct answer is '12'. Can you explain this answer?
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