Geometry Questions for SSC CPO PDF

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Geometry Questions for SSC CPO PDF
Geometry Questions for SSC CPO PDF

Geometry Questions for SSC CPO PDF

Download Top-15 SSC CPO Geometry Questions and Answers PDF, based on asked questions in previous CPO & other SSC exam papers.

Download Geometry Questions for SSC CPO PDF

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Question 1: What is the total number of circles passing through the two fixed points?

a) 1

b) 2

c) 4

d) Infinite

Question 2: Find the volume $(in cm^3)$ of a cube of side 4.5 cm.

a) 55.467

b) 14.445

c) 91.125

d) 26.465

Question 3: If the sum of the interior angles of a regular polygon is $720^\circ$ then how many sides does it have?

a) 8

b) 9

c) 6

d) 10

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Question 4: The length of one side and the diagonal of a rectangle are 20 cm and 29 cm respectively. Find the length of its other side (in cm).

a) 42

b) 30

c) 21

d) 60

Question 5: AB is the chord of circle of length 6 cm. From the center of the circle a perpendicular is drawn which intersects the chord at M and distance between centre and chord is 4 cm. find the area $(in cm^2)$ of the circle)

a) 55

b) 61.5

c) 70

d) 78.5

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Question 6: The two chords AB and CD of a circle intersect at point P, such that BP=4 cm, PD=5 cm, and CP=8 cm. find the length of chord AB.

a) 8

b) 10

c) 12

d) 14

Question 7: Find the total surface area $(in cm^2)$ of a right circular cylinder of diameter 28 cm and height 12 cm.

a) 2200

b) 2080

c) 1920

d) 2288

Question 8: The circumference of a circle is 110 cm. Find its radius (in cm).

a) 35

b) 19.5

c) 17.5

d) 39

Question 9: The area of a square is $30.25 cm^2$. Find its perimeter (in cm).

a) 44

b) 23

c) 22

d) 46

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Question 10: The base and height of a right angled triangle is 12 cm and 5 cm respectively. Find the circum-radius (in cm) of the triangle.

a) 5

b) 6

c) 6.5

d) 7

Question 11: For the circle shown below, find the length (in cm) of the largest cord of the circle.

a) 8

b) 12

c) 16

d) 18

Question 12: The volume of a cube is $274.625 cm^3$. Find its side (in cm).

a) 7.5

b) 6.5

c) 5.5

d) 3.5

Question 13: If the perimeter of a semicircle is 72 cm, then find its area $(in  cm^2)$.

a) 308

b) 616

c) 160

d) 320

Question 14: The area of a square is $42.25 cm^2$. Find its perimeter (in cm).

a) 52

b) 26

c) 28

d) 56

Question 15: The area of a right angled triangle ABC, right angled at B, is 46 sq units. A median is drawn from A to BC which intersects at D. Find the area (in sq. units) of triangle ABD.

a) 12

b) 23

c) 46

d) 88

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Answers & Solutions:

1) Answer (D)

Let there be 2 fixed points, as we can see from the figure that there can be infinite circles that passes through these 2 fixed points.

=> Ans – (D)

2) Answer (C)

Side of cube = $a=4.5$ cm

=> Volume of cube = $a^3$

= $(4.5)^3=91.125$ $cm^3$

=> Ans – (C)

3) Answer (C)

Sum of all interior angles of a polygon with $’n’$ sides = $(n-2)\times180^\circ$

Let the number of sides be $n$

=> Sum of interior angles = $(n-2)\times180^\circ=720^\circ$

= $(n-2)=\frac{720^\circ}{180^\circ}=4$

=> $n=4+2=6$

=> Ans – (C)

4) Answer (C)

Let the length of rectangle = $l$ cm and breadth, $b=20$ cm

=> Diagonal, $d^2=l^2+b^2$

=> $l^2=(29)^2-(20)^2$

=> $l^2=841-400=441$

=> $l=\sqrt{441}=21$

$\therefore$ Length of its other side = 21 cm

=> Ans – (C)

5) Answer (D)

Given : AB = 6 cm and OM = 4 cm

To find : Area of circle = ?

Solution : Let $r$ be the radius of circle

Also, MB = $\frac{6}{2}=3$ cm

In right $\triangle$ MOB,

=> $(OB)^2=(OM)^2+(MB)^2$

=> $(OB)^2=(4)^2+(3)^2$

=> $r^2=16+9=25$

=> $r=\sqrt{25}=5$ cm

$\therefore$ Area of circle = $\pi r^2$

= $3.14 \times(5)^2=78.5$ $cm^2$

=> Ans – (D)

6) Answer (D)

Given : AB and CD are chords of the circle which intersect at point P. BP = 4 cm, CP = 8 cm and PD = 5 cm

To find : AB = ?

Solution : Let AP = $x$ cm

Now, when two chords intersect each other inside a circle, the products of their segments are equal.

=> $(AP)\times(BP)=(CP)\times(DP)$

=> $x\times4=8\times5$

=> $x=\frac{40}{4}=10$

$\therefore$ AB = AP + PB = $10+4=14$ cm

=> Ans – (D)

7) Answer (D)

Height of cylinder, $h=12$ cm and radius, $r=\frac{28}{2}=14$ cm

Total surface area of cylinder = $2\pi r(r+h)$

= $2\times\frac{22}{7}\times14\times(14+12)$

= $88\times26=2288$ $cm^2$

=> Ans – (D)

8) Answer (C)

Let radius of circle = $r$ cm

=> Circumference = $2\pi r=110$

=> $2\times\frac{22}{7}\times r=110$

=> $r=110\times\frac{7}{44}$

=> $r=2.5\times7=17.5$ cm

=> Ans – (C)

9) Answer (C)

Let side of square = $s$ cm

=> Area = $s^2=30.25$

=> $s=\sqrt{30.25}=5.5$ cm

$\therefore$ Perimeter = $4\times5.5=22$ cm

=> Ans – (C)

10) Answer (C)

Length of base and height of triangle = 12 and 5 cm

=> Let length of hypotenuse = $x$ cm

=> $(x)^2=(12)^2+(5)^2$

=> $(x)^2=144+25=169$

=> $x=\sqrt{169}=13$ cm

Also, the circumcentre of a right angled triangle lies on its hypotenuse, thus circumradius = $\frac{1}{2}\times$ (hypotenuse)

= $\frac{1}{2}\times13=6.5$ cm

=> Ans – (C)

11) Answer (C)

Given : AT is tangent on the circle. AT = 6 cm and AB = 10 cm

To find : Largest chord = Diameter = ?

Solution : In right $\triangle$ ABT

=> $(BT)^2=(AB)^2-(AT)^2$

=> $(BT)^2=(10)^2-(6)^2$

=> $(BT)^2=100-36=64$

=> $BT=\sqrt{64}=8$ cm

$\therefore$ Diameter = $2\times8=16$ cm

=> Ans – (C)

12) Answer (B)

Let side of cube = $a$ cm

Volume of cube = $a^3=274.625$

=> $a=\sqrt[3]{6.5\times42.25}$

=> $a=6.5$ cm

=> Ans – (B)

13) Answer (A)

Let radius of semi circle = $r$ cm

=> Perimeter of semi circle = $\pi r+2r=72$

=> $r(\frac{22}{7}+2)=72$

=> $r(\frac{22+14}{7})=72$

=> $r=72\times\frac{7}{36}=14$ cm

$\therefore$ Area of semi-circle = $\frac{1}{2} \pi r^2$

= $\frac{1}{2}\times\frac{22}{7}\times(14)^2$

= $22\times14=308$ $cm^2$

=> Ans – (A)

14) Answer (B)

Let side of square = $s$ cm

=> Area = $s^2=42.25$

=> $s=\sqrt{42.25}=6.5$ cm

$\therefore$ Perimeter of square = $4s=4\times6.5=26$ cm

=> Ans – (B)

15) Answer (B)

Note :- A median divides a triangle into two parts of equal areas.

Proof :

It is given that $ar(\triangle ABC)=46$ sq.units

Also, AD bisects BC, let BC = $2x$ units => BD = $\frac{2x}{2}=x$ units

$\therefore$ $\frac{ar(\triangle ABD)}{ar(\triangle ABC)}=\frac{\frac{1}{2}\times(AB)\times(BD)}{\frac{1}{2}\times(AB)\times(BC)}$

=> $\frac{\triangle}{46}=\frac{x}{2x}$

=> $\triangle=\frac{46}{2}=23$ sq.units

=> Ans – (B)

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