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The length and breadth of a rectangle are both prime numbers, and its perimeter is 40 cm. Then the maximum possible area of the rectangle (in $$\text{cm}^2$$) is
Correct Answer: 91
Half the perimeter is $$\frac{40}{2} = 20$$ cm, so the length and the breadth are two primes adding up to 20. The possible prime pairs are 3 and 17, giving an area of $$3 \times 17 = 51$$, and 7 and 13, giving an area of $$7 \times 13 = 91$$. The greater of the two is $$91\ \text{cm}^2$$.
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