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CAT 2024 Slot 3 — QA Question 2

Geometric CentersEasy

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 âˆ†XYZ is

Answer & solution

Answer: 90

Solution

Medium

The points X, Y, Z are where each median meets the midpoint-triangle's sides. By the geometry of the medial triangle, each median bisects the side of the medial triangle it crosses, so X, Y, Z are themselves midpoints. Track the area through successive midpoint triangles, each of which has one-quarter the area of its parent.

1

Medial triangle. M, N, P are midpoints of the sides of △ABC\triangle ABC, so △MNP\triangle MNP (the medial triangle) has one-quarter the area of △ABC\triangle ABC.

[△MNP]=14[△ABC]⇒ [△MNP]=14(1440)(given area)⇒ [△MNP]=360\begin{aligned} &[\triangle MNP]=\tfrac14[\triangle ABC]\\ &\Rightarrow\ [\triangle MNP]=\tfrac14(1440)\quad\text{(given area)}\\ &\Rightarrow\ [\triangle MNP]=360 \end{aligned}
2

X, Y, Z are midpoints of MNP's sides. The median from AA passes through the centroid and the midpoint of BCBC; the segment MPMP is the midline parallel to BCBC. A median of △ABC\triangle ABC meets the corresponding midline at its midpoint. Hence X,Y,ZX, Y, Z are the midpoints of MP,MN,NPMP, MN, NP, making △XYZ\triangle XYZ the medial triangle of △MNP\triangle MNP.

[△XYZ]=14[△MNP]⇒ [△XYZ]=14(360)(from step 1)⇒ [△XYZ]=90\begin{aligned} &[\triangle XYZ]=\tfrac14[\triangle MNP]\\ &\Rightarrow\ [\triangle XYZ]=\tfrac14(360)\quad\text{(from step 1)}\\ &\Rightarrow\ [\triangle XYZ]=90 \end{aligned}
[△XYZ]=90 sq cm[\triangle XYZ]=90\text{ sq cm}

Two nested medial triangles scale area by 14×14=116\tfrac14\times\tfrac14=\tfrac1{16}. So [△XYZ]=144016=90[\triangle XYZ]=\dfrac{1440}{16}=90.

CAT 2024 Slot 3 QA Q2: The midpoints of sides AB, BC, and AC in ∆ABC are M, N, and P, respectively. The medians drawn from A, B, an — Solution | TheCATExam