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
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)
We hope this Geometry Questions PDF for SSC CPO Exam will be highly useful for your preparation.
