Surface area and volume solved problem for class IX

Q. The external and internal diameters of a hollow hemi-spherical vessel are 16cm and 12cm respectively the cost of painting 1 sq. cm of surface is Rs.2 .find the cost of painting the vessel?

Given, R = 16 cm and r = 12 cm

Area of outer surface = 2πR ²

Area of inner surface = 2πr ²

Area of the ring at the top = πR ² – πr ²

∴ Total area to be painted = (2πR ² + 2πr ² )+ (πR ² – πr ²) = π(3R ² + r ²)

= 22/7 x(3 x 16 x 16 + 12 x 12)=2866.29 sq. cm

Cost of painting = Rs (2 × 2866.29) = Rs 5732.57

Q. If a cone I cut into two parts by a horizontal plane passing through the mind – point if its axis. Find the ratio of the volumes of the upper part and the cone.

Let the height and radius of the given cone be H and R respectively.

The cone is divided into two parts by drawing a plane through the mid points of its axis and parallel to the base. One part is a smaller cone and the other part is a frustum of cone.

Þ OC = CA = H/2

Let the radius of smaller cone be r cm.

In ΔOCD and ΔOAB,

∠OCD = ∠OAB = 90°

∠COD = ∠AOB (common)

∴ ΔOCD ∼ ΔOAB (AA similarity criterion)

ÞOA/OC =AB/CD=OB/OD Þ H/H/2 =R/r Þ  R=2r

Now, Volume of smaller cone / Volume of cone  =( 1/3 X p CD² X OC) / (1/3 X p AB² X OA )

  = (1/3 X p r² X H/2) ¸  (1/3 X p R² X H ) Þ (1/3 X p r² X H/2) ¸  {1/3 X p (2r)² X H }

=1/6 ¸4/3 =1:8    ∴ Ratio of volume of upper part to the cone is 1:8.

Q. A solid cylinder is melted and cast into a cone of same radius .Find  the ratio of their  heights ?

Let the height of the cylinder = h₁

Let the height of the cone = h₂

Volume of the cylinder (V₁) = πr²h₁

Volume of the cone formed (V₂)= (1/3) πr²h₂

Since cone is formed from the cylinder, hence the volume is same.

Therefore,

(V₁) = (V₂)

πr²h₁ = (1/3) πr²h₂

3 h₁ = h₂

h₁/ h₂ = 1/3

h₂ / h₁ = 3/1

Therefore, height of the cone and cylinder are in the ratio of 3 : 1.

CBSE 8th Mensuration Solved Questions

11. A flooring tile has a shape of a parallelogram whose base is 28cm and the corresponding height is 20cm . how many such tiles are required to cover a floor of a area 2800cm²

Solution: Area of each parallelogram tile

= Base of the parallelogram × corresponding height of the parallelogram

= 28 cm × 20 cm= 560 cm²

Area of the floor = 28 m² = 2800 × (100 cm)² = 2800 × 10000 cm²

Number of tiles × Area of each parallelogram tile = Area of the floor

Number of tiles = Area of the floor/ Area of each parallelogram tile

                            = 2800 × 10000 cm²  /560 cm²   = 50000 tiles

12. The rainfall recorded on a certain day was  5cm. Find the volume of water that fell on 2 hectare field.

Solution: The volume of water= Area of ground x h= 2 hectare x 5cm = 2x10000m² x 5/100m=1000m³ = 1000x1000Lit=1000000Lit.{1m3=1000L}

13. Rain water which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm. What will be the height of water in the cylindrical vessel if the rain fall is 1 cm  (Take π = 3.14)

Solution : Let the height of the water level in the cylindrical vessel be h cm

Volume of the rain water = 600 × 400 × 1 cm³

Volume of water in the cylindrical vessel = π (20)2 × h cm³

A/Q,   600 × 400 × 1 = π (20)2 × h  or h =600/3.14 cm = 191 cm

14. The areas of three adjacent faces of a cuboid are 180 cm sq., 96 cm sq. and 120cm sq. what the volume of cuboid is. Please give me the answer

Solution:  lx b = 180       b x h = 96        h x l =120

l x b x b x h x h x  l = 180 x 96 x 120

(lbh)²= 180 x 96 x120

l b h = √(180 x 96 x120 ) =2073600

Volume   = √2073600   = 1440 cm³

15.  If a solid cylinder has a total surface area 462 sq. cm. & CSA  is 1/3rd of it so what is the volume of the cylinder?

Solution : Let the radius and height of the cylinder be r cm and h cm respectively.

Given, Total surface area of the cylinder = 462 cm²

∴ 2p r (r + h) = 462 cm²       .. .(1)

Lateral surface area of cylinder =  1/3 × Total surface area of cylinder (Given)

∴ 2prh =1/3 × 462 = 154 cm² ... (2)

From (1) and (2), we have

{2pr(r+h}/{2pr} = 462 cm²       / 154 cm²

(r+h) / h = 3

r = 3h-h=2h

r = 2h

From (2), we have

2prh =154 cm²

(2 x22rxr) /(7x2)}

r = 7cm

h = r/2=7/2=3.5cm

∴ Volume of the cylinder  = pr²h =22x7x7/3.5=539 cm²

 CBSE ADDA: 8th Mensuration Solved Questions: {For question 1 to 10 }

CBSE Practice Test Paper Volume and Surface area

Practice problems for class Mathematics

1. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold? 

2. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density  of the metal is 8.9 g per cm3 ? 

3. The diameter of the Moon is approximately one fourth the diameter of the Earth. What fraction is the volume of the Moon of the volume of the Earth? 

4. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m, then find the volume of the iron used to make the tank. 

5. Find the volume of a sphere whose surface area is 154 cm2. 

6. A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of Rs 498.96. If the rate of white wash is Rs 2.00 per square  meter, find   (i) inside surface area of the dome, (ii) the volume of the air inside the dome. 

7. A hemispherical bowl of steel is of thickness 0.3 cm . If the inner radius of the bowl is 7 cm, find the volume of the steel used in making the bowl. 

8. The volume of a conical tent is 1232 m3 and the area of its base is 154 m2.Find the length of the canvas required to build the tent, if the canvas is 2 m in width.

9. A cylindrical bucket, 14 cm in radius, is filled with water to some height. If a  rectangular solid of size 28 cm ×11 cm × 10 cm is immersed in the water, find theheight by which water rises in the bucket. 

10. A 11 cm × 4 cm rectangular piece of paper is folded and taped without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder so formed.

11. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? 

12. The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find (i) height of the cone,(ii) slant height of the cone,(iii) curved surface area of     the cone.  

13. A conical tent is 9 m high with base diameter 24 m. Find the number of persons it  can accommodate if each person requires

(i) 2 m2 space on the ground,

(ii) 15 m3space to breathe,

(iii) 2 m2 space on the round and 15 m3 space to breathe. 

14. A right triangle ABC with its sides 5 cm, 12 cm and13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained  

15.If the triangle ABC with its sides 5 cm, 12 cm and13 cm is revolved about the side 5 cm, then find the   volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtaine

8th Volume surface are solved examples

8th Volume surface are solved Examples Download File
For Practice: Test Paper-VOLUMES SURFACE AND AREA--VIII
1. Find the area of a carpet spread on a floor of 12 m by 9.5 m   
2.A tiles measure 10 cm x 10 cm How many tiles are required for a wall 4 m by 2.5 m
3.A lane is 45 m by 32 m. Find cost of leveling it at Rs. 2.50 per m2?
4. A card board is 4.5 m by 3.5 m. Find the cost of painting it on both side at Rs1.25 per m2
5.A field is 350 m by 300 m. Find the length of wire needed for fencing
6.A foot path runs inside the rectangular 38 m and 32 m. If the width of path is runs inside .The width of path is 5m. Find the area of path?
7.A side of square is 300 m .Find the perimeter of square in km.
8. Find the perimeter of the rectangular field 725 m by 300 m in km.
9 . The radius of the circle is 3 m. What is the circumference of another circle, whose area is 49 times that of the first?
10. Two circles touch externally. The sum of their areas is 130 p sq. cm and the distance between their centers is 14 cm. Find the radii of the circles.
11. A wire when bent in the form of an equilateral triangle encloses an area of 121 √3 cm2 . If the same wire is bent in the form of a circle, find the area of the circle.

8th Compound Interest Solved problems

Ex.1. Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually.

  Sol.  Amount = Rs [7500*(1+(4/100)2] = Rs (7500 * (26/25) * (26/25)) = Rs. 8112.            

            Therefore, C.I. = Rs. (8112 - 7500) = Rs. 612.

Ex. 2. Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually.

   Sol.         Time = 2  years  4  months = 2(4/12) years = 2(1/3) years.
                    Amount = Rs'. [8000 X (1+­(15/100))2 X (1+((1/3)*15)/100)]

                         =Rs. [8000 * (23/20) * (23/20) * (21/20)]     = Rs. 11109.                             .

                   :. C.I. = Rs. (11109 - 8000) = Rs. 3109.

    Ex. 3. Find the compound interest on Rs. 10,000 in 2 years at 4% per annum, the   interest being compounded half-yearly

IX CBSE Guess Paper: Ch: Surface area and Volume

IX Chapter - 13 : Surface areas and Volumes

1. A solid cylinder has a total surface area of 231cm2. Its curved surface area is 2/3 of the total surface area. Find the volume of the cylinder.

Ans: 269.5cm2

2. The diameter of a garden roller is 1.4m and it is 2m long. How much area will it cover in 5 revolutions?

Ans: 44m2

3. Three metal cubes whose edge measure 3cm, 4cm and 5cm respectively are melted to form a single cube, find its edge.

Ans: 6cm

4. The dimensions of a cubiod are in the ratio of 1 : 2 : 3 and its total surface area is 88m2. Find the dimensions.

Ans: 2, 4, 6 cm

5. Find the lateral curved surface area of a cylindrical petrol storage tank that is 4.2m in diameter and 4.5m high. How much steel was actually used, if 1/12  of steel actually used was wasted in making the closed tank.

Ans: 59.4m2, 95.04m2

MCQ : VIII Surface Area and Volume

MCQ : VIII Surface Area and Volume Maths Help For Class 8th(VIII)

1. Lateral surface  area  of  a  cylinder

(a) πr²h   b)  2πrh    c)  2πr[r+h]

2.Total  surface  area  of  a  cylinder

(a) 2πrh  b)  2πr²h  c)  2πr[r+h]

3. When  the  radius  is  doubled  surface  area  of  a  cylinder  increases  by ___  times .

(a)  2   b)  4  c)  6  d) 8

4.When  the  height  is  doubled  the  lateral  surface  area  of  a  cylinder  increases  by  ----  times .

(a) 2  b)  4  c)  6  d)  8

5.There  are  2  cuboid  boxes  having  measurements  60x40x50  and  50x50x50  .Which  box  requires  the  lesser  amount  of  material  to  make  .

(a) 60  x  40 x 50  b)  50x50x50  c)  both same .

6.Surface  area  of  a  cube  having  side  6cm

(a)  6  b)  6²  c)  6³  d)  none of  these

7.Total  surface  area  of  a  cuboid  includes  the  area  of

(a)  4 faces  b)  2 faces  c)  6 faces  d)  3 faces

8.Ratio  between  the  lateral  area  and  base  area  of  a cuboid

(a) 1:2  b)  2: 1  c)  4:1  d)  1:4

9. Lateral  surface   area  of  a  cuboid

(a)  Base  area  x  height  b)  Base  perimeter  x  height  c)  volume/2

10.  Two  cubes  each  with  side  b  are  joined  to  form  a  cuboid . What  is  the  surface  area  of  this  Cuboid

(a) 12b²  b)10b²  c) 18b²  d) none of these .

11. Base  perimeter  of  a  room  is 34m  and  the  height  of  the  room  is  10m . Find  the  area  of  the  four  side  walls .

(a) 34x10 m²  b) 2x34x10 m² c)34x10x10 m²  d)none of these.

12. The  curved  surface  area  of  a  cylinder  is  100πm². Length  of  the  cylinder  is  10 m . Find  its  Radius .

(a) 10m  b) 5m  c) 20m d) none  of  these

Solution:

1)  b   2) c   3) b   4) a  5) c   6) c  7) c  8) c   9) b   10) b   11) a 12) b

CBSE TEST PAPERS Download

CBSE I NCERT 10th Arithmetic Progressions : Problems for self evaluation.

10th Ch- A. P Problems for self evaluation. Download File

10th Quadratic Equation Solved Question and Self Evaluation Question

10th CBSE Maths Chapter: Quadratic Equation Solved Question and Self Evaluation Question

More Download Paper  Click on link  Self Evaluation Question

Quadratic Equation Solved Question and Self Evaluation Question part-1
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Quadratic Equation Solved Question and Self Evaluation Question part-2
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Assignment Class X Quadratic Equations
1. Find the value of k for kx2 + 2x - 1 = 0, so that it has two equal roots

2. Find the value of k for k x2 - 2√ 5 x + 4 = 0, so that it has two equal roots.

3. If the roots of the equation (b - c) x2 + (c - c) x + (a - b) = 0 are equal, accordingly prove that 2b = a + c.

4. Find the discriminant of the quadratic equation 3x2– 4 √3 x + 4 = 0, and hence find the nature of its roots.

5. Find the value of k for 2 x2 + k x + 3 = 0, so that it has two equal roots.

6. Find the value of k for k x (x – 2) + 6 = 0, so that it has two equal roots.

7. Find the value of k for which the equation x2 + 5kx + 16 = 0 has no real roots.

8 Find the discriminant of the quadratic equation 2x2– 6x + 3 = 0, and hence find the nature of its roots.

9. Find the value of k for k2 x2 – 2 (2 k - 1) x + 4 = 0, so that it has two equal roots.

10. Find the value of k for (k + 1) x2 – 2 ( k - 1) x + 1= 0, so that it has two equal roots.