Maths Mensuration Questions Answers With Hindi Explanation
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Maths Mensuration Questions Answers With Hindi Explanation
Maths Mensuration Questions Answers With Hindi Explanation
Q1.There is a pyramid on a base which is a regular hexagon of side 2a cm. If every slant edge of this pyramid is of length 5a/2 cm, then the volume of this pyramid is
(a) 3a3 cm3
(b) 3 √2a3 cm3
(c) 3√3a3 cm3
(d) 6a3 cm3
Option: (c)
Explaination :
Area of the base =6×√3/4 ×(2a)2 =6×√3/4 ×4a2=6√3a2 sq.cm
Height
Volume of pyramid=1/3× area of the base × height=1/3×6√3a2 ×3/2 a=3√3a3 cm3
Q2.The length of a diagonal of a square is 15√2 cm. Its area is
(a) 112.5 cm2
(b) 450 cm2
(c) 225√2/2 cm2
(d) 225 cm2
Option: (d)
Explaination :Diagonal of square = √2×side
√2×side=15√2
⇒side=15√2 /√2=15
Area of square =15×15=225sq.cm
Q3.The respective heights and volumes of a hemisphere and a right circular cylinder are equal, then the ratio of their radii is:
(a)√2 : √3
(b)√3 : 1
(c)√3 : √2
(d) 2 :√3
Option: (c)
Explaination :
Radius of hemisphere = height of cylinder = r units
Q4.A toy is in the form of a cone mounted on a hemisphere. The radius of the hemisphere and that of the cone is 3 cm and height of the cone is 4 cm. The total surface area of the toy is
(a) 75.43 sq cm
(b) 103.71 sq cm
(c) 85.35 sq cm
(d) 120.71 sq cm
Option: (b)
Explaination :
Total surface area of the toy =πrl+πr2
Q5. At each corner of a triangular field of sides 26m, 28m and 30m, a cow is tethered by a rope of length 7m. The area (in m2) ungrazed by the cows is:
(a) 366
(b) 259
(c) 154
(d) 77
Option: (b)
Explaination :Area grazed by all cows = 180º/360º πr2
πr2 /2 =1/2×22/7×7×7 =77 sq.meter
Semi-perimeter of triangular field(s) =26+28+30/2 =42 meter
Area of the field =
Area ungrazed by the cows = 336 – 77 = 259sq.metre
Q6.The volumes of two spheres are in the ratio 8 : 27. The ratio of their surface areas is:
(a) 4 : 9
(b) 2 : 3
(c) 4 : 5
(d) 5 : 6
Option: (a)
Explaination :
Let the volumes be 8x3 and 27x3
Their radius are 2x and 3x
The ratio of their surface area= 4x2 : 9x2 =4 :9
Q7.The height of the cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume is 1/27 of the volume of the cone, at what height, above the base, is the section made?
(a) 6 cm
(b) 8 cm
(c) 10 cm
(d) 20 cm
Option: (d)
Explaination :Let H and R be the height and radius of bigger cone respectively and h and r that of smaller cone.
From triangles AOB and AMN, A is common and MN ? OB. Triangles AOB and AMN are similar,
so AO/AM =BO/MN
⇒30/h=R/r..............(i)
Volume of smaller cone =1/3πr2h
Volume of bigger cone =1/3πR2H
According to the question, 1/3πr2h=(1/3πR2H)×1/27
⇒r2h=R2H/27
27r2h=R2H
27h/H =R2/r2
27h/H =(30/h)2
27h/H =900/h2⇒ 27h3=900H=900×30
h3=900×30/27 =1000
h=10cm
Required height = 30 – 10 = 20 cm
Q8.A sphere and a cylinder have equal volume and equal radius. The ratio of the curved surface area of the cylinder to that of the sphere is:
(a) 4 : 3
(b) 2 : 3
(c) 3 : 2
(d) 3 : 4
Option: (b)
Explaination :
Q9.The diameter of the base of a right circular cone is 4 cm and its height 2√3 cm. The slant height of the cone is:
(a) 5 cm
(b) 4 cm
(c) 2√3 cm
(d) 3 cm
Option: (b)
Explaination :Slant height of cone
Q10.The ratio of the length of the parallel sides of a trapezium is 3 : 2. The shortest distance between them is 15 cm. if the area of the trapezium is 450 cm2, the sum of the lengths of the parallel sides is:
(a) 15 cm
(b) 36 cm
(c) 42 cm
(d) 60 cm
Option: (d)
Explaination :Area of the trapezium = 1/2 (Sum of parallel sides)×altitude
450=1/2(3x+2x)×15
5x=450×2/15
=60cm
Q11.An equilateral triangle and a regular hexagon have the same perimeter. The ratio of the area of the triangle to that of the hexagon is:
(a) 3 : 2
(b) 2 : 3
(c) 1 : 2
(d) 1 : 4
Option: (b)
Explaination :
If the side of an equilateral triangle be x units and side of regular hexagon be y units,
then 3x=6y ⇒x/y=2
√3/4x2 /6×√3/4 y2
1/6×4=2/3 ;
2:3
Q12.In measuring the sides of a rectangle, there is an excess of 5% on one side and 2% deficit on the other. Then the error percent in the area is:
(a) 3.3
(b) 3.0
(c) 2.9
(d) 2.7
Option: (c)
Explaination :Required percentage = (x+y+xy/100)%
Negative sign for decrease =(5-2-10/100)%=2.9%
Q13.A sphere and a cube have equal surface areas. The ratio of the volume of the sphere to that of the cube is:
(a) √π:√6
(b) √6:√π
(c) √2 :√π
(d)√π:√3
Option: (b)
Explaination :
If the radius of the sphere be r units and the edge of the cube be x units, then
Surface area of sphere = Surface area of cube ⇒ 4πr2=6x2
r2/x2 =6/4π
3/2π; r/x=√3/√2π
Q14.A circle and a square have equal areas. The ratio of a side of the square and the radius of the circle is
(a) 1 : √π
(b) √π:1
(c) 1 :π
(d) π : 1
Option: (b)
Explaination :
If the radius of circle be r units and the side of square be x units,
then X2 =πr2
⇒x2/r2=π/1
x/r=√π/1
Q15.Surface areas of three adjacent faces of a cuboid are p, q, r. its volume is
(a)
(b)
(c)
(d)(√pqr)
Option: (d)
Explaination :
If the length, breadth and height of a cuboid be t, b and h units respectively,
then P = lb, q = bh, r = hl
pqr=l2b2h2
Volume of the cuboid = lbh =√pqr
Q16.A copper wire, when bent in the form of a square encloses a region having area 121 cm2. If the same wire is bent in the form of a circle, the area of the region enclosed by the wire will be [π = 22/7]
(a) 154 cm2
(b) 143 cm2
(c) 132 cm2
(d) 121 cm2
Option: (a)
Explaination :Side of a square = √area =√121=11cm
Length of wire = Perimeter of square= 4 ×11=44cm
Circumference of circle =2rπ
2rπ=44
r=7cm
Area of circle-πr2=22/7×7×7 =154 cm2
Q17.A solid cone of height 9 cm with diameter of its base 18 cm is cut out from a wooden solid sphere of radius 9 cm. The percentage of wood wasted is:
(a) 25
(b) 30
(c) 50
(d) 75
Option: (d)
Explaination :Volume of sphere =
4/3 πr3 =4/3π ×9×9×9 =972π cu. cm.
Volume of cone =1/3πR2H
1/3π ×9×9×9 =243πcu.cm.
Percentage of wood wasted =(972π-243π)/972π ×100
=75%
Q18.The area of an equilateral triangle is 4√3 cm2. The length of each side of the triangle is:
(a) 3 cm
(b) 2√2 cm
(c) 2√3 cm
(d) 4 cm
Option: (d)
Explaination :Area of the equilateral triangle = √3/4 ×side2
4√3=√3/4 ×side2
side2=4√3×4/√3 =16
side=√16=4cm
Q19.The radius of a circle is increased by 1%. How much does the area of the circle increase?
(a) 1%
(b) 1.1%
(c) 2%
(d) 2.01%
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Option: (d)
Explaination :Effect on area = (1+1+11/100)% =2.01%
Q20.The circumference of a circle is 100 cm. The measure of a side of the square inscribed in this circle is
(a) 25√2πcm
(b) 50√2/π cm
(c) 50√2π cm
(d) 25√2/π cm
Option: (b)
Explaination :Circumference of the circle = π×Diameter=100
d=100/π cm
Q21.In a cylindrical vessel of diameter 24 cm filled up with sufficient quantity of water, a solid spherical ball of radius 6 cm is completely immersed.The increase in height of water level is:
(a) 1.5 cm
(b) 2 cm
(c) 3 cm
(d) 4.2 cm
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Option: (b)
Explaination :If the height of increased water level be h cm, then
πr2h=4/3 πR3
12×12×h=4/3×6×6×6
h=2 cm
Q22.A cow is tied on the corner of a rectangular field of size 30m ×20m by a 14m long rope. The area of the region, that she can graze, is :(π = 22/7)
(a) 350 m2
(b) 196 m2
(c) 154 m2
(d) 22 m2
Option: (c)
Explaination :area of the region, that cow can graze
90/360 π r2
90/360 ×22/7 ×14×14 =154 sq.meter
Q23.The diameters of two cylinders, whose volumes are equal, are in the ratio 3 : 2. Their heights will be in the ratio:
(a) 4 : 9
(b) 5 : 6
(c) 5 : 8
(d) 8 : 9
Option: (a)
Explaination :
Q24.A 7 m wide road runs outside around a circular park, whose circumference is 176 m. The area of the road is: [π = 22/7]
(a) 1386 m2
(b) 1472 m2
(c) 1512 m2
(d) 1760 m2
Option: (a)
Explaination :
If the radius of the circular park be r metre, then,
2πr=176
2×22/7×r=176
r=28 meter
Radius of the park with road = 28 + 7 = 35 metre
Area of the road =22/7(352-282)=22/7×63×7=1386 meter2
Q25.A cube and a sphere have equal surface areas. The ratio of their volumes is:
(a) π : 6
(b) √π : √6
(c) √6 : √π
(d) 6 :π
Option: (b)
Explaination :
Surface area of cube = 6?2 ;
Surface area of sphere = 4πr2 ;
6?2 = 4πr2
⇒l/r=2√π/√6