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

Entered answer:

✅ Correct Answer: 90

We have triangle ABC with area = 1440 sq cm. M, N, P are midpoints of sides AB, BC, AC respectively.

Solution figure for CAT 2024 QA question 11 (Geometry)

The medians of triangle ABC intersect triangle MNP at points X, Y, Z.

The trickest part of this question is getting the graph right. Post which, the solution is pretty simple.

When you connect the midpoints of a triangle's sides, you get the medial triangle.

The medial triangle has area = $\tfrac{1}{4}$ × (original triangle's area)

The medial triangle is similar to the original triangle with scale factor $\tfrac{1}{2}$ . Since area scales as (scale factor) $^2$ , we get $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$ .

Area of triangle MNP = $\tfrac{1}{4} \times 1440 = 360$ sq cm

When the medians of a triangle intersect the sides of its medial triangle, the resulting triangle has area equal to $\tfrac{1}{16}$ of the original triangle's area. XYZ is nothing but the triangle from the median of MNP (hence, $\tfrac{1}{4}$ of MNP). It's like a shrunk up triangle of ABC.

This is a specific case that comes from the properties of medians and how they interact with medial triangles.

Area of triangle XYZ = $\tfrac{1}{16} \times$ Area of triangle ABC

Area of triangle XYZ = $\tfrac{1}{16} \times 1440 = 90$ sq cm

Therefore, the area of triangle XYZ is 90 sq cm.

CAT 2024 Slot 2

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CAT 2020 Slot 2

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CAT 2017 Slot 2

Let $P$ be an interior point of a right-angled isosceles triangle $ABC$ with hypotenuse $AB$ . If the perpendicular distance of P from each of $A B, B C$ , and $C A$ is $4(\sqrt{2}-1) \mathrm{cm}$ , then the area, in sq cm, of the triangle $A B C$ is

2025 Slot 3

A triangle ABC is formed with AB = AC = 50 cm and BC = 80 cm. Then, the sum of the lengths, in cm, of all three altitudes of the triangle ABC is

CAT 2017 Slot 1

Let $ABC$ be a right-angled isosceles triangle with hypotenuse $BC$ . Let $BQC$ be a semi-circle, away from $A$ , with diameter $BC$ . Let $BPC$ be an arc of a circle centred at $A$ and lying between $BC$ and $BQC$ . If $AB$ has length $6$ cm ,then the area, in sq. cm, of the region enclosed by $BPC$ and $BQC$ is:

CAT 2023 Slot 1

In a right-angled triangle $\triangle A B C$ , the altitude $A B$ is $5 \mathrm{~cm}$ , and the base $B C$ is $12 \mathrm{~cm}$ . $P$ and $Q$ are two points on $B C$ such that the areas of $\triangle \mathrm{ABP}, \triangle \mathrm{ABQ}$ and $\triangle \mathrm{ABC}$ are in arithmetic progression. If the area of $\triangle \mathrm{ABC}$ is $1.5$ times the area of $\triangle \mathrm{ABP}$ , the length of PQ in cm , is

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