Single correct option
ABCD is a square of side 10 cm. E is the midpoint of side BC. The line segment AE intersects the diagonal BD at F. What is the area of triangle BEF (in cm²)?
- A. 10
- B. 50/3
- C. 25/4
- D. 25/3
Show answer and solution
Answer: (D) 25/3
Place A(0, 0), B(10, 0), C(10, 10), D(0, 10). Then E = (10, 5).
Line AE: y = x/2. Diagonal BD: x + y = 10. Solving: x + x/2 = 10 ⇒ x = 20/3, y = 10/3. So F = (20/3, 10/3).
Triangle BEF has base BE = 5 lying on the line x = 10; its height is the horizontal distance of F from that line = 10 − 20/3 = 10/3.
Area = ½ × 5 × 10/3 = 25/3 cm².
Alternative: Triangles FBE and FDA are similar (BE ∥ AD) with ratio BE : AD = 1 : 2, so BF : FD = 1 : 2, and F is 1/3 of the way along BD from B. Area(ABE) = 25 and F divides AE in 2 : 1, so area(BEF) = 25/3.
Line AE: y = x/2. Diagonal BD: x + y = 10. Solving: x + x/2 = 10 ⇒ x = 20/3, y = 10/3. So F = (20/3, 10/3).
Triangle BEF has base BE = 5 lying on the line x = 10; its height is the horizontal distance of F from that line = 10 − 20/3 = 10/3.
Area = ½ × 5 × 10/3 = 25/3 cm².
Alternative: Triangles FBE and FDA are similar (BE ∥ AD) with ratio BE : AD = 1 : 2, so BF : FD = 1 : 2, and F is 1/3 of the way along BD from B. Area(ABE) = 25 and F divides AE in 2 : 1, so area(BEF) = 25/3.
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