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
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
Updated On: Jan 13, 2026
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Correct Answer: 90
Solution and Explanation
Given:
- \( M, N, P \) are midpoints of sides \( AB, BC, \) and \( AC \) respectively in \( \triangle ABC \).
- The medians from vertices \( A, B, C \) intersect line segments \( MP, MN, \) and \( NP \) at points \( X, Y, Z \) respectively.
- Area of \( \triangle ABC = 1440 \, \text{cm}^2 \)
Concept:
The triangle \( \triangle XYZ \) formed by intersections of medians with the sides of the medial triangle has area equal to \( \frac{1}{16} \) of the area of triangle \( \triangle ABC \).
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: \( \boxed{90 \, \text{cm}^2} \)
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