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.

Q. if the height of a cylinder is doubled, by what number must the radius of the base be multiplied so that the resulting cylinder has the same volume as the original cylinder?

V = p R²h

If the height is doubled the new volume would be

v = pr²(2h) = 2pr²h

If v =V then

p R²h =  2pr²h

R²  = 2r²

 R²/2  = r²

√R²/2  = r

1/√2 x R  = r

R would have to be multiplied by 1 / √2

v = 2p (1/2R)²h = pr²h = Original volume V.

 Q. If each edge of a cube is increased by 50% then what is the percentage increase in the surface area of the cube?

Initital Surface Area Of Cube = 6a²

Final Edge Of Cube = a+50/100*x = 3a/2

FInal Surface Area = 6x3a/2x 3a/2 = 54a²/4 = 27a²/2

Increase In Surface Area = 27a²/2 - 6a² = 15a²/2

Now Percentage = Increase/Initial x 100

15a²/2 / 6a² x  100 = 15a²/2 x 1/6a² x  100 = 125%

Q. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm.

Given, height of conical tent = 8 m

and radius of tent = 6 m

Now, slant height of tent l = √ h² + r² = √ 8² + 6²  = 10m

∴ Curved surface area of tent = p rl= .14 x 6 x 10=188.4 sq. m 

Given, breadth of tarpaulin sheet = 3 m

Let the length of tarpaulin sheet required will be equal to x m.

It is given that 20 cm (or 0.2 m) length of material is wasted in stitching margins and cutting.

∴ Effective length of tarpaulin sheet = (x - 0.2) m

It is quite clear that, area of sheet = curved surface area of tent

⇒(x- 0.2) × 3 = 188.4

⇒(x- 0.2) = 62.8

⇒x = 62.8 + 0.2 = 63 m

Hence, the tarpaulin sheet of 63 m length will be required to make conical height of given dimensions.

Q.If four times the sum of the areas of two circular faces of a cylinder of height 8cm is equal to the twice the curved surface area then find the diameter of the cylinder.

Let the radius of the given cylinder be 'r' cm and height be 'h' cm.

Given, height of cylinder = h = 8 cm

Now, area of circular face of cylinder = pr² 

Therefore, area of two circular faces =pr²   + pr²   = 2pr² 

Also, curved surface area of cylinder = 2pr x 8 =16pr 

According to the given condition,

4 × area of two circular faces of cylinder = 2 × curved surface area of cylinder

4x 2pr²  = 2 x 16pr 

r²      =  4r

r = 4   So, r = 4 cm Hence, diameter of the cylinder = 2 × 4 cm = 8 cm