# CAT Practice : Coordinate Geometry

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Geometry - Triangles

## Basic Geometry

Q.44:
The Olympics committee came up with a new rule. The flag of the gold medal winning team would be hoisted to the right (AB) at 5m. The flag of silver medal winning team would be hoisted to the left (PQ) at a height of 3m. The flag (MN) of bronze medal winning team would be hoisted at the point of intersection of the line joining the top of each of AB and PQ to the foot of other, as shown in the figure above. A and P are 8m apart. In a wrestling event, India won the bronze medal. Find the height at which the Indian flag was hoisted.
1. 2 m
2. ${\frac{5}{2}}$ m
3. ${\frac{5}{8}}$ m
4. ${\frac{15}{8}}$ m

Choice (D). ${\frac{15}{8}}$ m

## Detailed Solution

ΔPAB ~ ΔPMN
Therefore, $\frac{PM}{PA} = \frac{MN}{AB}$
=) $\frac{PM}{8} = \frac{MN}{5}$…………………(1)
Also, ΔAMN~ΔAPQ
Therefore, $\frac{AM}{AP} = \frac{MN}{PQ}$
=) $\frac{AM}{8} = \frac{MN}{3}$…………………(2)
$\frac{(PM + AM)}{8} = \frac{MN}{5} + \frac{MN}{3}$
=) $\frac{8}{8} = \frac{(3MN + 5MN)}{15}$ (As, PM + AM = AP)
=) $MN = \frac{15}{8}$

Correct Answer: ${\frac{15}{8}}$ m

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Geometry is probably the most vital topic as far as CAT preparation is concerned. Geometry sets the stage for Trigonometry, Cogeo and Mensuration as well.