Area of triangle formed by intersection of medians and midpoint segments

CAT 2024 Slot 3 · QA · Hard · Geometry

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

Answer

90

Explanation

In ABC\triangle ABC, M,N,PM, N, P are midpoints of AB,BC,ACAB, BC, AC. Thus MNP\triangle MNP is the medial triangle of ABC\triangle ABC, and its area is 14Area(ABC)=360\frac{1}{4} \text{Area}(\triangle ABC) = 360 sq cm.

Median from vertex AA passes through midpoint NN of BCBC. Since line segment MPMP joins the midpoints of ABAB and ACAC, MPBCMP \parallel BC and MPMP bisects ANAN. Thus, the intersection XX of median ANAN with MPMP is the midpoint of MPMP.

Similarly, YY is the midpoint of MNMN and ZZ is the midpoint of NPNP.

Hence, XYZ\triangle XYZ is the medial triangle of MNP\triangle MNP. Area(XYZ)=14Area(MNP)=116Area(ABC)=144016=90 sq cm\text{Area}(\triangle XYZ) = \frac{1}{4} \text{Area}(\triangle MNP) = \frac{1}{16} \text{Area}(\triangle ABC) = \frac{1440}{16} = 90\text{ sq cm}

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