# CAT Practice : Coordinate Geometry

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Geometry - Triangles

## Basic Geometry

Q.42: M and N are two points on the side PQ and PR of a triangle PQR respectively such that MNQR is a trapezium and MN:QR = 2:5. Find the ratio of the area of triangle PMN : Trapezium MNQR.
1. 4:25
2. 1:2
3. 4:21
4. 4:10

Choice (C). 4:21

## Detailed Solution

Given, MNQR is a trapezium =) MN || QR
Therefore, ΔPMN ~ ΔPQR
Since, MN:QR = 2:5, heights of ΔPMN and ΔPQR will also be in the ration 2:5.
=) $\frac{AΔPMN}{A MNQR} = \frac{AΔPMN}{( AΔPQR – AΔPMN)} = \frac{\frac{1}{2} * 2b * 2h}{\frac{1}{2} * 5b * 5h - \frac{1}{2} * 2b * 2h}$ = $\frac{4}{(25-4)} = \frac{4}{21}$

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