The diagonal of a rectangle is 10 cms and is twice the length of one of the sides. What is the area of the rectangle in sq. cm? (R.R.B. 2006)

### Concept & Right Triangle Properties This problem requires using a known diagonal and its proportional relationship to one side to find the second side via the Pythagorean theorem. $$ \text{Area} = \text{Length} \times \text{Breadth} $$ ### Step-by-Step Solution 1. We are given the diagonal ($d$) $= 10 \text{ cm}$. 2. The diagonal is twice the length of one of the sides. Let this known side be $w$ (width). $$ d = 2w $$ $$ 10 = 2w \implies w = 5 \text{ cm} $$ 3. Let the other side (length) be $l$. Use the Pythagorean theorem to find $l$: $$ d^2 = l^2 + w^2 $$ $$ 10^2 = l^2 + 5^2 $$ $$ 100 = l^2 + 25 $$ 4. Isolate $l^2$: $$ l^2 = 100 - 25 = 75 $$ 5. Take the square root to find $l$: $$ l = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \text{ cm} $$ 6. Calculate the area of the rectangle: $$ \text{Area} = l \times w $$ $$ \text{Area} = 5\sqrt{3} \times 5 = 25\sqrt{3} \text{ sq. cm} $$ ### Exam Strategy & Shortcut If a right triangle has a hypotenuse that is exactly twice the length of one leg, it is a special $30^\circ-60^\circ-90^\circ$ triangle. The sides of such a triangle are in the ratio $1 : \sqrt{3} : 2$. Since the hypotenuse is $10$ (representing the $2$ part), the shorter side is $5$ (the $1$ part), and the longer side is $5\sqrt{3}$ (the $\sqrt{3}$ part). Area $= 5 \times 5\sqrt{3} = 25\sqrt{3}$. ### Common Pitfall A common mistake is stopping after finding the unknown side ($5\sqrt{3}$) and assuming this is the final answer, neglecting the final step of multiplying the sides together to find the area. ### Final Answer Therefore, the correct answer is **$25\sqrt{3}$**.