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Mensuration Questions for SSC-CPO Set-2 PDF

Download SSC CPO Mensuration Questions with answers set-2 PDF based on previous papers very useful for SSC CPO exams. Very important Mensuration Questions for SSC exams.

Question 1: The circumference of a circle is 88 cm. Find its radius (in cm).

a) 28

b) 15

c) 14

d) 30

Question 2: The volume of a hemisphere is 19404 cm3. Find its diameter (in cm).

a) 42

b) 21

c) 84

d) 63

Question 3: If the perimeter of a semicircle is 36 cm, then find its area (in $cm^{2}$)

a) 154

b) 35

c) 77

d) 70

Question 4: The lengths of the two diagonals of a rhombus are 7 cm and 24 cm. Find the length of its perimeter (in cm).

a) 25

b) 100

c) 75

d) 50

Question 5: Find the total surface area (in $cm^{2}$) of a right circular cone of diameter 21 cm and slant height 11 cm.

a) 467.5

b) 384

c) 724

d) 709.5

Question 6: Calculate the area (in $cm^{2}$) of a circle of radius 17.5 cm.

a) 1925

b) 962.5

c) 809.5

d) 1619

Question 7: The area of an equilateral triangle is 25√3 $cm^{2}$. Find its side (in cm).

a) 10

b) 5

c) 20

d) 30

Question 8: Find the total surface area $(in cm^2)$ of a cuboid of length, breadth and height of 10.5 cm, 8 cm and 9 cm respectively.

a) 607

b) 767

c) 769

d) 501

Question 9: The length of the diagonal and the breadth of a rectangle is 26 cm and 10 cm respectively. Calculate its area $(in cm^2)$

a) 480

b) 96

c) 240

d) 192

Question 10: Find the total surface area (in cm²) of a right circular cylinder of diameter 42 cm and height 14 cm.

a) 4488

b) 4250

c) 4010

d) 4620

Let radius of circle = $r$ cm

=> Circumference = $2\pi r=88$

=> $2\times\frac{22}{7}\times r=88$

=> $r=88\times\frac{7}{44}$

=> $r=2\times7=14$ cm

=> Ans – (C)

Let radius of hemisphere = $r$ cm

Volume of hemisphere = $\frac{2}{3}\times\pi r^3$

=> $\frac{2}{3}\times\frac{22}{7}\times(r)^3=19404$

=> $(r)^3=19404\times\frac{21}{44}$

=> $(r)^3=441\times21$

=> $r=\sqrt[3]{21\times21\times21}=21$ cm

$\therefore$ Diameter = $2\times21=42$ cm

=> Ans – (A)

Let radius of semi circle = $r$ cm

=> Perimeter of semi circle = $\pi r+2r=36$

=> $r(\frac{22}{7}+2)=36$

=> $r(\frac{22+14}{7})=36$

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

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

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

= $11\times7=77$ $cm^2$

=> Ans – (C)

Given : ABCD is a rhombus and AC = 24 cm and BD = 7 cm

To find : Perimeter of ABCD

Solution : Diagonals of a rhombus bisect each other at right angle.

=> BE = $\frac{7}{2}=3.5$ cm and AE = $\frac{24}{2}=12$ cm

Thus, in right $\triangle$ AEB,

=> $(AB)^2=(AE)^2+(BE)^2$

=> $(AB)^2=(12)^2+(3.5)^2$

=> $(AB)^2=144+12.25=156.25$

=> $AB=\sqrt{156.25}=12.5$ cm

$\therefore$ Perimeter of rhombus ABCD = $4\times12.5=50$ cm

=> Ans – (D)

Radius of cone, $r=\frac{21}{2}=10.5$ cm and slant height, $l=11$ cm

Total surface area of cone = $\pi r(l+r)$

= $\frac{22}{7}\times10.5\times(10.5+11)$

= $33\times21.5=709.5$ $cm^2$

=> Ans – (D)

Radius of circle = $r=17.5$ cm

=> Area = $\pi r^2$

= $\frac{22}{7}\times17.5\times17.5$

= $22\times2.5\times17.5=962.5$ $cm^2$

=> Ans – (B)

Let side of equilateral triangle = $s$ cm

=> Area = $\frac{\sqrt3}{4} s^2=25\sqrt3$

=> $\frac{s^2}{4}=25$

=> $s^2=25\times4=100$

=> $s=\sqrt{100}=10$ cm

=> Ans – (A)

Length, $l=10.5$ cm, breadth, $b=8$ cm and height, $h=9$ cm

Total surface area of cuboid = $2(lb+bh+hl)$

= $2[(10.5\times8)+(8\times9)+(9\times10.5)]$

= $2\times(84+72+94.5)$

= $2\times250.5=501$ $cm^2$

=> Ans – (D)

Let the length of rectangle = $l$ cm and breadth, $b=10$ cm

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

=> $l^2=(26)^2-(10)^2$

=> $l^2=676-100=576$

=> $l=\sqrt{576}=24$ cm

$\therefore$ Area = $24\times10=240$ $cm^2$

=> Ans – (C)

Height of cylinder, $h=14$ cm and radius, $r=\frac{42}{2}=21$ cm
Total surface area of cylinder = $2\pi r(r+h)$
= $2\times\frac{22}{7}\times21\times(21+14)$
= $132\times35=4620$ $cm^2$