Let $ABCD$ be a rectangle inscribed in a circle of radius $13 \mathrm{~cm}$. Which one of the following pairs can represent, in cm , the possible length and breadth of $ABCD$ ?

When a rectangle is inscribed in a circle, all four vertices of the rectangle lie on the circle. This creates a special relationship:

The diagonal of the rectangle becomes the diameter of the circle.

When you connect opposite vertices of a rectangle, you get the diagonal. Since all vertices lie on the circle, this diagonal passes through the center and becomes the diameter. This is because the rectangle's opposite vertices are the farthest apart, making the diagonal the longest possible chord - which is the diameter.

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Given:

• Circle radius = 13 cm
• Therefore, diameter = 2 × 13 = 26 cm
• Rectangle has length = $l$ and breadth = $b$

Since diagonal of rectangle = diameter of circle:

$\text{Diagonal} = 26 \text{ cm}$

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For any rectangle, the diagonal, length, and breadth form a right triangle:

$\text{Diagonal}^2 = \text{Length}^2 + \text{Breadth}^2$

Substituting our values:

$26^2 = l^2 + b^2$

$676 = l^2 + b^2$

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We need to find integer values of $l$ and $b$ such that $l^2 + b^2 = 676$.

Let's try $l = 24$ and $b = 10$:

$24^2 + 10^2 = 576 + 100 = 676$

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The pair $(24, 10)$ represents a valid rectangle because both dimensions are positive, they satisfy our constraint equation, and the resulting rectangle can physically fit inside the circle.

Therefore, the possible length and breadth of rectangle $ABCD$ are 24 cm and 10 cm respectively.

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For any rectangle inscribed in a circle, the diagonal equals the diameter. This transforms geometry problems into algebra using the Pythagorean theorem!