Most Exclusive Questions on Geometry for SSC CGL Tier-II Exam

By Neha Uppal|Updated : August 22nd, 2019

Most Exclusive Questions on Geometry for SSC CGL Tier-II ExamAs you know, SSC CGL Tier-II exam is going to be held across India from 11-13th September. You all must be busy in your preparation for the D-Day. Geometry is one of the most imp topics which is very important from the examination's perspective. You can expect around 10-12 questions from this topic in your exam. To help you in your preparation, we have selected some exclusive questions on Geometry that will act as an icing on the cake. We have also provided a detailed solution below each question to help you with the method or approach. 

Most Exclusive Questions on Geometry for SSC CGL Tier-II Exam

1. In ∆ABC, E is a point such that BE is an angle bisector of B & AE  BE. F is a point on AC such that EF || BC. If AC = 12 cm, find the length of AF.

A. 8 cm
B. 4 cm
C. 9 cm
D. 6 cm

Answer ||| D

Solution |||

We know in an isosceles triangle ABC

4

If AD ⊥BC then, AD is angle bisector of A or vice versa

∴ here since BE is angle bisector and BE ⊥ AD.

5

∴ ∆ABD is an isosceles ∆ where AB = BD

Also BE is bisecting AD at E

∴ By Thales theorem, EF || CD

∴ F is also the mid-point of AC

Hence, length of AF byjusexamprep cm

2.In the given figure byjusexamprep and byjusexamprep are the centers of two circles of same radius R. A third circle of center byjusexamprep and radius byjusexamprepis tangent to both circles. If distance between byjusexamprep and byjusexamprep is thrice of radius of smaller circle, find the distance between byjusexamprep and the line byjusexamprep

byjusexamprep

 

A. byjusexamprep 
B. byjusexamprep 
C. byjusexamprep 
D. byjusexamprep 

Answer ||| A

Solution |||

byjusexamprep

byjusexamprep
byjusexamprep
byjusexamprep
byjusexamprep
byjusexamprep
byjusexamprep
byjusexamprep
byjusexamprep

3.In the given figure there is byjusexamprep ABC . D is a point on BC. byjusexamprepBAD = 30byjusexamprep and byjusexamprepCAD = 60byjusexamprep. Find the length of side AD .

byjusexamprep

 

A. byjusexamprep
B. byjusexamprep
C. byjusexamprep
D. byjusexamprep

Answer ||| A

Solution |||

Area of triangle = byjusexamprepAB.AC sinbyjusexamprep . Where byjusexamprep is angle between AB and AC.

Area of triangle byjusexamprep ABC = Area of triangle byjusexamprepABD + Area of triangle byjusexamprepDAC

byjusexamprep

byjusexamprep

AD = byjusexamprep

4.ABCD is a quadrilateral which circumscribed a circle of radius 7 cm. AB||CD, byjusexamprep and byjusexamprep. What is the perimeter of quadrilateral?

A. byjusexamprep 
B. byjusexamprep 
C. byjusexamprep 
D. byjusexamprep 

Answer ||| A

Solution |||

byjusexamprep

byjusexamprep

OB is the angle bisector of angle B.

byjusexamprep
byjusexamprep
byjusexamprep

Again,

byjusexamprep
byjusexamprep
byjusexamprep

Perimeter

byjusexamprep
byjusexamprep
byjusexamprep
byjusexamprep

 

5.ABCD is a rectangle. P is a point inside the rectangle such that AP = 3 cm, PC = byjusexamprep cm and DP = 5 cm. If byjusexamprep what is the length AB of the rectangle?

A. 7 cm
B. byjusexamprep cm
C. byjusexamprep
D. byjusexamprep cm

Answer ||| C

Solution |||

byjusexamprep

byjusexamprep
byjusexamprep
byjusexamprep
byjusexamprep

Since APB is a right triangle,

byjusexamprep
byjusexamprep
byjusexamprep

6.There are two identical circles of radius 4 cm each. If the length of the direct common tangent is 14 cm, then what is the length of the transverse common tangent?

A. byjusexamprep 
B. byjusexamprep 
C. byjusexamprep 
D. byjusexamprep 

Answer ||| B

Solution |||

byjusexamprep

Since, both circles are identical, AB = OM = 14 cm. and OC = CM = 14/2 = 7 cm. Now. OPC and CQM are right triangles and congruent also.

byjusexamprep
byjusexamprep
byjusexamprep

7.In the given figure ABC is a right triangle with byjusexamprep and byjusexamprep If the radius of the larger circle is 6 cm, what is the radius of smaller circle?

byjusexamprep

 

A. 4 cm 
B. 3 cm 
C. 1 cm 
D. 2 cm 

Answer ||| D

Solution |||

byjusexamprep

OC will be angle bisector of C.

byjusexamprep
byjusexamprep
byjusexamprep

Now,

byjusexamprep
byjusexamprep
byjusexamprep

 

8.What is the ratio of the inradius, circumradius and one of the ex-radii of an equilateral triangle?

A. 1:2:3 
B. 1:2:5 
C. 2:3:5 
D. 1:2:4 

Answer ||| A

Solution |||

Property: Ratio between the inradius, circumradius and ex-radius is 1:2:3.

 

9.ABC is a triangle such that AB = AC. BD and CE are the medians to the sides AC and AB respectively. If BDCE, then value of AB/BC =?

A. byjusexamprep
B. 2/√3
C. 4/3
D. 3/2

Answer ||| A

Solution |||

byjusexamprep

ABC is an isosceles triangle, BE = CD=AB/2 = AC/2 and BD = EC and hence BO = OC, OE = OD. Also, ED=BC/2 Now,

byjusexamprep
byjusexamprep
byjusexamprep
byjusexamprep
2 BE2 = 5 BC2/4
2AB2/4 = 5BC2/4
AB/BC = byjusexamprep

10.In the given figure PA=10,RC=X and QB = 15, Find the value of X.(byjusexamprepPAB=byjusexamprepQBC=byjusexamprepRCB=90byjusexamprep)

byjusexamprep

 

A. 6
B. 10
C. 8
D. 12 

Answer ||| A

Solution |||

In byjusexamprep BAP and byjusexamprep BCP

byjusexamprep ------- (1)

In byjusexamprep ARC and byjusexamprep AQB

byjusexamprep -------- (2)

Adding equation (1) and (2)

byjusexamprep

 

byjusexamprep

 

10byjusexamprep15 = 10x + 15x

x = byjusexamprep

 

11.If two opposite vertex in a rectangle are (2,1) and (5,3) and the equation of other diagonal is x-2y+k=0. find k?

A. 1/2
B. 3/2
C. 2
D. 6

Answer ||| A

Solution |||

M is the mid-point of diagonal AC & BD .

Untitled.png

Co-ordinate of M = byjusexamprep

line x-2y+k passes through M(7/2,2)

hence byjusexamprep

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