# Geometry Questions for CAT

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Geometry Questions for CAT:

You can download the Geometry questions or you can go through the details below.

Question 1:

A one rupee coin is placed on a table. The maximum number of similar one rupee coins which can be placed on the table, around it, with each one of them touching it and only two others is

A. 8

B. 6

C. 10

D. 4

Question 2:

Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer then how many such triangles exist?

A.
5

B.
10

C.
21

D.
15

Question 3:

Each side of a given polygon is parallel to either the X or the Y axis. A corner of such a polygon is said to be convex if the internal angle is 90° or concave if the internal angle is 270°. If the number of convex corners in such a polygon is 25, the number of concave corners must be

A. 20

B. 0

C. 21

D. 22

Question 4:

A farmer has decided to build a wire fence along one straight side of this property. For this, he planned to place several fence-posts at six metre intervals, with posts fixed at both ends of the side. After he bought the posts and wire, he found that the number of posts he had bought was five less than required. However, he discovered that the number of posts he had bought would be just sufficient if he spaced them eight metres apart. What is the length of the side of his property and how many posts did he buy?

A.
20

B.
0

C. 21

D.
22

Question 5:

In a triangle ABC, the lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD = 3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC?

A. 17.05

B. 27.85

C. 22.45

D. 26.25

If we join centres of 2 outer circles with the centre circle, it will make an equilateral triangle.
Hence at the centre it will make an angle of $$60^o$$ so total $$\frac{360^o}{60^o} = 6$$ triangles will be there
Hence, 6 outer circles will be there.

For obtuse-angles triangle, $$c^2 > a^2 + b^2$$ and c < a+b
If 15 is the greatest side, 8+x > 15 => x > 7 and $$225 > 64 + x^2$$ => $$x^2$$ < 161 => x <= 12
So, x = 8, 9, 10, 11, 12
If x is the greatest side, then 8 + 15 > x => x < 23
$$x^2 > 225 + 64 = 289$$ => x > 17
So, x = 18, 19, 20, 21, 22
So, the number of possibilities is 10

Let the total number of sides be x.
(25*90)+(x-25)*270 = (x-2)180
x = 46
Number concave corners = x-25 = 46-25 = 21

Let the number of posts the farmer bought = x
Length of the side = (x+5-1)*6 = 6*(x+4)
Length of the side is also equal to (x-1)*8
=> 8x – 8 = 6x + 24 => x = 16
Length of side = 6*20 = 120 m