# Quantitative Aptitude – Geometry – Mensuration

## Slot – 1 – Quantitative Aptitude – Geometry – Mensuration – AB is a diameter of a circle of radius 5 cm

Q. AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to?
1. 9.3
2. 8.5
3. 9.1
4. 7.8

Solution: See the figure, Given AB is diameter = 5*2 = 10 , BP =6 And AQ = AP/2
Angle P and Q will be right angled (angle in semi circle is always equals to 90 degree)
Thus in right angled triangle APB , AP = (AB^2 – BP^2)^1/2 = (10^2 – 6^2 )^1/2 = 8
So AQ = 8/2 = 4
Now in right angled triangle AQB, BQ = (AB^2 – AQ^2)^1/2 = (10^2 – 4^2 )^1/2 = (84)^1/2 = 9.1 approx

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