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# SSC CGL Geometry Questions

Download Top-20 Geometry questions for SSC CGL exam. Most importantÂ  Geometry questions based on asked questions in previous exam papers for SSC CGL.

Question 1:Â A tangent is drawn to a circle of radius 6 cm from a point situated at a distance of 10 cm from the centre of the circle. The length of the tangent will be

a)Â 7 cm

b)Â 4 cm

c)Â 5 cm

d)Â 8 cm

Question 2:Â How many perfect squares lie between 120 and 300 ?

a)Â 5

b)Â 6

c)Â 7

d)Â 8

Question 3:Â A copper wire of length 36 m and diameter 2 mm is melted to form a sphere. The radius of the sphere (in cm) is:

a)Â 2.5

b)Â 3

c)Â 3.5

d)Â 4

Question 4:Â The product of two numbers is 45 and their difference is 4. The sum of squares of the two numbers is

a)Â 135

b)Â 240

c)Â 73

d)Â 106

Question 5:Â The square root ofÂ $14 + 6\sqrt{5}$

a)Â $2 + \sqrt{5}$

b)Â $3 + \sqrt{5}$

c)Â $5 + \sqrt{3}$

d)Â $3 + 2\sqrt{5}$

Question 6:Â A copper wire is bent in the form of an equilateral triangle, and has an area $121\sqrt{3}$ cm2Â . If the same wire is bent into the form of a circle, the area(in cm2) enclosed by the wire in(Take $\pi = \frac{22}{7}$)

a)Â 364.5

b)Â 693.5

c)Â 346.5

d)Â 639.5

Question 7:Â The square root of 0.09 is

a)Â 0.30

b)Â 0.03

c)Â 0.81

d)Â 0.081

Question 8:Â If an obtuse-angled triangle ABC, is the obtuse angle and O is the orthocenter. If = 54$^{\circ}$ , then is

a)Â 108$^{\circ}$

b)Â 126$^{\circ}$

c)Â 136$^{\circ}$

d)Â 116$^{\circ}$

Question 9:Â If S is the circumcentre of and = 50$^{\circ}$ , then the value of is

a)Â 20$^{\circ}$

b)Â 40$^{\circ}$

c)Â 60$^{\circ}$

d)Â 80$^{\circ}$

Question 10:Â AC and BC are two equal cords of a circle. BA is produced to any point P and CP, when joined cuts the circle at T. Then

a)Â CT : TP = AB : CA

b)Â CT : TP = CA : AB

c)Â CT : CB = CA : CP

d)Â CT : CB = CP : CA

Question 11:Â The external bisectors of and of meet at point P. If = 80$^{\circ}$ , the is

a)Â 50$^{\circ}$

b)Â 40$^{\circ}$

c)Â 80$^{\circ}$

d)Â 100$^{\circ}$

Question 12:Â The total surface area of a sphere is $8\pi$ square unit. The volume of the sphere is

a)Â $\frac{8}{3}\pi$

b)Â $8\sqrt{3}\pi$

c)Â $\frac{8\sqrt{3}}{5}\pi$

d)Â $\frac{8\sqrt{2}}{3}\pi$

Question 13:Â A conical flask is full of water. The flask has base radius $r$ and height $h$. This water is poured into a cylindrical flask of base radius mr. The height of water in the cylindrical flask is

a)Â $\frac{h}{2} m^2$

b)Â $\frac{2h}{m}$

c)Â $\frac{h}{3m^2}$

d)Â $\frac{m}{2h}$

Question 14:Â In a triangle ABC,Â $\angle{A} = 90^o , \angle{C} = 55^o , \overline{AD}$ is perpendicular to $\overline{BC}$. What is the value of $\angle{BAD}$ ?

a)Â 60 $^{\circ}$

b)Â 45 $^{\circ}$

c)Â 55 $^{\circ}$

d)Â 35 $^{\circ}$

Question 15:Â If G is the centroid of triangle ABC and area of triangle ABC = 48cm2, then the area of triangle BGC is

a)Â 8 cm2

b)Â 16 cm2

c)Â 24 cm2

d)Â 32 cm2

Question 16:Â The diagonals AC and BD of a cyclic quadrilateral ABCD intersect each other at the point P. Then, it is always true that

a)Â AP . CP = BP . DP

b)Â AP . BP = CP . DP

c)Â AP . CD = AB . CP

d)Â BP . AB = CD . CP

Question 17:Â A, B, C, D are four points on a circle. AC and BD intersect at a point such that $\angle{BEC} = 130^o$ and $\angle{ECD} = 20^o$. Then, $\angle{BAC}$ is

a)Â 90 $^{\circ}$

b)Â 100 $^{\circ}$

c)Â 110 $^{\circ}$

d)Â 120 $^{\circ}$

Question 18:Â In a triangle, if three altitudes are equal, then the triangle is

a)Â equilateral

b)Â right

c)Â isosceles

d)Â obtuse

Question 19:Â ABCD is a cyclic trapezium with AB || DC and AB = diameter of the circle. If angleCAB = $30^{\circ}$, then angleADC is

a)Â $60^{\circ}$

b)Â $120^{\circ}$

c)Â $150^{\circ}$

d)Â $30^{\circ}$

Question 20:Â ABC is a triangle. The bisectors of the internal angle $\angle$B and external $\angle$C intersect at D. If $\angle$BDC=$50^{\circ}$, then $\angle$A is

a)Â $100^{\circ}$

b)Â $90^{\circ}$

c)Â $120^{\circ}$

d)Â $60^{\circ}$

radius OA = 6 and OB = 10

Now, in $\triangle$OAB right angle at A

AB = $\sqrt{(OB)^2 – (OA)^2}$

= $\sqrt{100-36} = \sqrt{64}$

= 8 cm

let’s say n be a number whose perfect square lie between 120 and 300
hence 120<$n^{2}$<300
or $121\leq n^{2} \leq289$
or $11\leq n^{2} \leq17$

since we know volume will remain same while melting
$\pi r_{1}^{2}h= \frac{4}{3}\pi r_{2}^{3}$
where $r_{1}$ is radius of cylinderical wire and $r_{2}$ is radius of sphere and h is length of wire
putting values we will get $r_{2}$ = 3 cm.

As we know $(a-b)^{2}$ = $a^{2} + b^{2} – 2ab$
We assume that first number is a and second number is b hence ab = 45
and a – b = 4
after putiing values we will getÂ $a^{2} + b^{2}$ = 106

Given question can be written as $9+5+6\sqrt{5}$
or it will be square ofÂ $3+\sqrt{5}$

Area of equilateral triangle is $\frac{\sqrt{3}}{4} a^{2}$ where a is side of triangle
which is equals to $121{\sqrt{3}}$
or a = 22 and whole length of wire will be 66
from here when it is bend to make a circle, circumference will be $2\pi r$ = 66
r = 10.5
hence area of circle will be $\pi r^{2}$ = 346.5

0.09 = $\frac{9}{100}$
$\sqrt{\frac{9}{100}} = \frac{3}{10}$ = 0.3

It is given that AC = BC, also $\triangle$ PTB andÂ $\triangle$Â PAC are similar, we haveÂ :

$\frac{CA}{CP}=\frac{BT}{BP}$ —————-(i)

Also, we haveÂ $\angle$Â PBC =Â $\angle$Â BTC ($\because$Â $\angle$Â PBC = $\angle$ BAC = $\angle$Â BTC) andÂ $\angle$Â PCB =Â $\angle$Â BCT

=>Â $\triangle$Â PBC $\sim$ $\triangle$Â BTC

Thus,Â $\frac{CB}{BP}=\frac{CT}{BT}$

=>Â $\frac{BT}{BP}=\frac{CT}{CB}$ ————–(ii)

From equations (i) and (ii), we getÂ :

$\frac{CA}{CP}=\frac{CT}{CB}$

=> Ans – (C)

Let the radius of the sphere be ‘r’
Hence $4\pi r^2$ = $8\pi$
=> r = $\sqrt{2}$
Volume of the cube = $\frac{4}{3}\pi r^3$ = $\frac{4}{3}\pi \sqrt{2}^3$ = $\frac{8\sqrt{2}}{3}\pi$

Volume of water in flask = $\frac{\pi r^2 h}{3}$ (where h is height of water in conical flask)
Volume of water in cylinder = $\pi m^2r^2 h_1$ (where h_1 is height of water in cylinderical flask)
Hence nowÂ $\frac{\pi r^2 h}{3}$ =Â $\pi m^2r^2 h_1$
or $h_1 = \frac{h}{3m^2}$

$\angle A = 90^o$
$\angle C = 55^o$
$\angle B$ will be $180 – (90+55) = 35^o$
As AD is perpendicular to BC Hence $\angle BAD=180-(90+35)=55^o$

As we know area of triangle, with centroid as one of the vertices and remaining 2 triangle vertices, is $\frac{1}{3}$rd of the area of whole triangle.
Hence area will be $\frac{48}{3}$ = 16

As we know that a cyclic quadrilateral can be inscribed into a circle, Hence in triangle APB and in triangle CPD.

$\angle PAB = \angle PDC$ (same sector angles)

$\angle PCD = \angle PBA$ (same sector angles)

Hence third angle will also be equal and they will be similar triangles.

So $\frac{AP}{PD} = \frac{BP}{PC}$

Angle ABD will be equal to angle ACD = $20^o$ (same sector angles)
Angle BEC = $130^o$ so angle AED = $130^o$ (concurrent angles)
Now angle BEA will be $\frac{360-130-130}{2} = 50^o$
So angle EDC will be $180-(50+20) = 110^o$

It is property of equilateral triangle that length of all its altitutes are equal.

letÂ angle CDA = x

since AB is parallel to CD, angle ACD=30 and angle CAD=30

in triangle ACD,

sum of all three angles = 180

30 + 30 + x = 180

x = 120

so the answer is option B.

In $\triangle$ BDC,

=> $y+(180^\circ-2x+x)+50^\circ=180^\circ$

=> $y-x+50^\circ=0$

=> $y-x=-50^\circ$

In $\triangle$ ABC,

=> $2y+(180^\circ-2x)+\angle A=180^\circ$

=> $2(y-x)+\angle A=0$

=> $2(-50^\circ)+\angle A=0$

=> $\angle A=100^\circ$

=> Ans – (A)