# Geometry Questions for CAT with solutions (Set-2)

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Geometry is of the important topic for CAT. The number of questions asked from this topic is high in the exam.

Geometry Questions for CAT with solutions (Set-2):

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

Question 1:

In triangleABC, ∠B is a right angle, AC = 6 cm, and D is the mid-point of AC. The length of BD is

A. 4 cm
B. 6 cm
C. 3 cm
D. 3.5 cm

Question 2:

In ABC, points P, Q and R are the mid-points of sides AB, BC and CA respectively. If area of ABC is 20 sq. units, find the area of PQR.

A. 10 sq. units
B. 5√3 sq. units
C. 5 sq. units
D. None of these

Question 3:

In a rectangle, the difference between the sum of the adjacent sides and the diagonal is half the length of the longer side. What is the ratio of the shorter to the longer side?

A. √3 : 2
B. 1 : √3
C. 2 : 5
D. 3 : 4

Question 4:

Ten points are marked on a straight line and eleven points are marked on another straight line. How many triangles can be constructed with vertices from among the above points?

A. 495
B. 550
C. 1045
D. 2475

Question 5:

There is a circle of radius 1 cm. Each member of a sequence of regular polygons S1(n), n = 4,5,6,… , where n is the number of sides of the polygon, is circumscribing the circle; and each member of the sequence of regular polygons S2(n), n = 4,5,6…. where n is the number of sides of the polygon, is inscribed in the circle. Let L1(n) and L2(n) denote the perimeters of the corresponding polygons of S1(n) and S2(n). Then $$\frac{L1(13)+2\pi }{L2(17)}$$ is

A. greater than $$\frac{\pi}{4}$$ and less than 1
B. greater than 1 and less than 2
C. greater than 2
D. less than $$\frac{\pi}{4}$$

A circle is circumscribed to triangle ABC.
For this circle D will be centre of circle, and AD , DC , BD will radius of this circle.

As we know, the triangle joining midpoints of sides will divide it in 4 similar triangles of equal area.
So area will be = $$\frac{20}{4} = 5$$

Consider rectangle with shorter side as $$y$$, longer side as $$x$$ and diagonal as $$z$$.
So $$(x+y) – z = \frac{x}{2}$$
Where $$z^2=x^2 + y^2$$
After putting value of z and solving we can get ratio of y to x as 3:4

For a triangle to be formed, we need three points.
Case 1: Select 2 points on the line that has 10 points and 1 point on the line that ha 11 points.
This can be done in $$^{10}C_2$$*$$^{11}C_1$$ ways = 495 ways.
Case 2: Select 2 points on the line that has 11 points and 1 point on the line that ha 10 points.
This can be done in $$^{11}C_2$$*$$^{10}C_1$$ ways = 550 ways.
495 + 550 = 1045 ways.

The perimeter of the circle is equal to 2$$\pi$$.
The perimeter of the polygon inscribing the circle is always greater than the perimeter of the circle => L1(13) > 2$$\pi$$
The perimeter of the polygon inscribed in the circle is always less than the perimeter of the circle => L2(13) < 2$$\pi$$
=> $$\frac{L1(13)+2\pi }{L2(17)}$$ > 2