Question

The area of a rectangle and the square of its perimeter are in the ratio 1 : 25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio

• A. 1:4
• B. 2:9
• C. 1:3
• D. 3:8

Answer 1

The Correct Option is A

Step 1: Let Variables Represent Dimensions

Let the length be \( L \) and the breadth be \( B \).

Step 2: Use Given Relationship

We are given:

\[ \frac{\text{Area of rectangle}}{(\text{Perimeter})^2} = \frac{1}{25} \]

Area = \( L \times B \) and Perimeter = \( 2(L + B) \). So:

\[ \frac{LB}{(2(L + B))^2} = \frac{1}{25} \]

Step 3: Simplify the Equation

\[ \frac{LB}{4(L + B)^2} = \frac{1}{25} \]

\[ 25LB = 4(L + B)^2 \]

Expand the right-hand side:

\[ 25LB = 4(L^2 + B^2 + 2LB) = 4L^2 + 4B^2 + 8LB \]

Step 4: Rearranging Terms

Bring all terms to one side:

\[ 25LB - 8LB = 4L^2 + 4B^2 \quad \Rightarrow \quad 17LB = 4L^2 + 4B^2 \]

\[ L^2 + B^2 = \frac{17}{4}LB \]

Step 5: Use a Ratio Substitution

Let \( \frac{L}{B} = r \Rightarrow L = rB \). Substitute into the equation:

\[ (rB)^2 + B^2 = \frac{17}{4} \cdot rB \cdot B \]

\[ r^2B^2 + B^2 = \frac{17}{4}rB^2 \Rightarrow B^2(r^2 + 1) = \frac{17}{4}rB^2 \]

Cancel \( B^2 \) from both sides:

\[ r^2 + 1 = \frac{17}{4}r \]

\[ 4r^2 + 4 = 17r \Rightarrow 4r^2 - 17r + 4 = 0 \]

Step 6: Solve the Quadratic

\[ r = \frac{17 \pm \sqrt{289 - 64}}{8} = \frac{17 \pm \sqrt{225}}{8} = \frac{17 \pm 15}{8} \]

\[ r = \frac{32}{8} = 4 \quad \text{or} \quad r = \frac{2}{8} = \frac{1}{4} \]

Step 7: Final Ratio

Since \( B < L \), the valid ratio is:

\[ \frac{B}{L} = \frac{1}{4} \quad \Rightarrow \quad \boxed{1 : 4} \]