Question

\( \triangle ABC \sim \triangle DEF \) and their areas are respectively \(75 \text{ cm}^2\) and \(48 \text{ cm}^2\). If \(EF = 4 \text{ cm}\), then \(BC = \_\_\_\_\_ \text{ cm}\).

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

In similar triangles, area ratio always equals the square of side ratio. If the area ratio becomes a perfect square fraction like \( \frac{25}{16} \), the corresponding side ratio immediately becomes \( \frac{5}{4} \).

Answer 1

The Correct Option is D

Concept: When two triangles are similar, their corresponding sides are proportional and the ratio of their areas is equal to the square of the ratio of corresponding sides. If: \[ \triangle ABC \sim \triangle DEF \] then: \[ \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \left(\frac{BC}{EF}\right)^2 \] This theorem is extremely useful because it connects areas directly with side lengths.

Step 1: Write the given information carefully. Area of \( \triangle ABC \): \[ 75 \text{ cm}^2 \] Area of \( \triangle DEF \): \[ 48 \text{ cm}^2 \] Corresponding side: \[ EF = 4 \text{ cm} \] We need to find: \[ BC = ? \]

Step 2: Apply the area ratio property of similar triangles. \[ \frac{75}{48} = \left(\frac{BC}{4}\right)^2 \] Let: \[ BC = x \] Then: \[ \frac{75}{48} = \left(\frac{x}{4}\right)^2 \]

Step 3: Simplify the fraction. Both numerator and denominator are divisible by 3: \[ \frac{75}{48} = \frac{25}{16} \] So: \[ \frac{25}{16} = \left(\frac{x}{4}\right)^2 \]

Step 4: Take square root on both sides. \[ \sqrt{\frac{25}{16}} = \frac{x}{4} \] \[ \frac{5}{4} = \frac{x}{4} \]

Step 5: Solve for \(x\). Multiply both sides by 4: \[ x = 5 \] Therefore: \[ BC = 5 \text{ cm} \]

Step 6: Verification of answer. Check side ratio: \[ \frac{BC}{EF} = \frac{5}{4} \] Square of ratio: \[ \left(\frac{5}{4}\right)^2 = \frac{25}{16} \] Area ratio: \[ \frac{75}{48} = \frac{25}{16} \] Both are equal, so answer is correct. Final Answer: \[ \boxed{5 \text{ cm}} \]