Single correct option
ABCD is a square of side 10 cm. E is the midpoint of BC. The line segment AE meets the diagonal BD at F. What is the area of triangle BEF?
- A. 25/2 sq cm
- B. 50/3 sq cm
- C. 10 sq cm
- D. 25/3 sq cm
Show answer and solution
Answer: (D) 25/3 sq cm
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. Intersection: x + x/2 = 10 → F = (20/3, 10/3).
Triangle BEF has base BE = 5 (along x = 10) and height = horizontal distance of F from x = 10 = 10 − 20/3 = 10/3.
Area = ½ × 5 × 10/3 = 25/3 sq cm.
Alternative: triangles FEB and FAD are similar (AD ∥ BE) with ratio BE : AD = 1 : 2, so F divides AE in 2 : 1; area(ABE) = 25, and BEF is 1/3 of it = 25/3.
Line AE: y = x/2. Diagonal BD: x + y = 10. Intersection: x + x/2 = 10 → F = (20/3, 10/3).
Triangle BEF has base BE = 5 (along x = 10) and height = horizontal distance of F from x = 10 = 10 − 20/3 = 10/3.
Area = ½ × 5 × 10/3 = 25/3 sq cm.
Alternative: triangles FEB and FAD are similar (AD ∥ BE) with ratio BE : AD = 1 : 2, so F divides AE in 2 : 1; area(ABE) = 25, and BEF is 1/3 of it = 25/3.
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