Question

The midpoints of sides AB, BC, and AC in ∆ABC are M, N, and P, respectively. The medians drawn from A, B, and C intersect the line segments MP, MN and NP at X, Y, and Z, respectively. If the area of ∆ABC is 1440 sq cm, then the area, in sq cm, of ∆XYZ is

Answer 1

Correct Answer: 90

Given:

• \( M, N, P \) are midpoints of sides \( AB, BC, \) and \( AC \) respectively in \( \triangle ABC \).
• The medians from vertices \( A, B, C \) intersect segments \( MP, MN, \) and \( NP \) at \( X, Y, Z \) respectively.
• Area of \( \triangle ABC = 1440 \, \text{cm}^2 \)

Concept:

The triangle \( \triangle XYZ \) formed by the intersection of medians with the sides of the medial triangle has an area equal to \( \frac{1}{16} \) of the area of \( \triangle ABC \).

Calculation:

\[ \text{Area of } \triangle XYZ = \frac{1}{16} \times \text{Area of } \triangle ABC = \frac{1}{16} \times 1440 = 90 \, \text{cm}^2 \]

Final Answer: \( \boxed{90 \, \text{cm}^2} \)