Area of triangle formed by intersection of medians and midpoint segments

Question

The midpoints of sides AB, BC, and AC in $	riangle 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 $	riangle ABC$ is 1440 sq cm, then the area, in sq cm, of $	riangle XYZ$ is

Accepted Answer

90. In $	riangle ABC$, $M, N, P$ are midpoints of $AB, BC, AC$. Thus $	riangle MNP$ is the medial triangle of $	riangle ABC$, and its area is $\frac{1}{4} 	ext{Area}(	riangle ABC) = 360$ sq cm. Median from vertex $A$ passes through midpoint $N$ of $BC$. Since line segment $MP$ joins the midpoints of $AB$ and $AC$, $MP \parallel BC$ and $MP$ bisects $AN$. Thus, the intersection $X$ of median $AN$ with $MP$ is the midpoint of $MP$. Similarly, $Y$ is the midpoint of $MN$ and $Z$ is the midpoint of $NP$. Hence, $	riangle XYZ$ is the medial triangle of $	riangle MNP$. $$	ext{Area}(	riangle XYZ) = \frac{1}{4} 	ext{Area}(	riangle MNP) = \frac{1}{16} 	ext{Area}(	riangle ABC) = \frac{1440}{16} = 90	ext{ sq cm}$$

Answer

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Area of triangle formed by intersection of medians and midpoint segments

Explanation

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