very important questions on surface area and volume class10

                SURFACE AREA AND VOLUME IMPORTANT NOTES

The concept of surface area and volume for Class 10 is provided here. In this article, we are going to discuss the surface area and volume for different solid shapes such as the cube, cuboid, cone, cylinder, and so on. The surface area can be generally classified into Lateral Surface Area (LSA), Total Surface Area (TSA), and Curved Surface Area (CSA). Here, let us discuss the surface area formulas and volume formulas for different three-dimensional shapes in detail. In this chapter, the combination of different solid shapes can be studied. Also, the procedure to find the volume and its surface area in detail.

Cuboid and its Surface Area

The surface area of a cuboid is equal to the sum of the areas of its six rectangular faces. Consider a cuboid whose dimensions are $l\times b\times \mathrm{h,}$ respectively.

 

Surf

Cuboid with length l, breadth b and height h

The total surface area of the cuboid (TSA) = Sum of the areas of all its six faces
TSA (cuboid) = $2(l\times b)+2(b\times h)+2(l\times h)=2(lb+bh+lh)$

Lateral surface area (LSA) is the area of all the sides apart from the top and bottom faces.
The lateral surface area of the cuboid = Area of face AEHD + Area of face BFGC + Area of face ABFE + Area of face DHGC
LSA (cuboid) = $2(b\times h)+2(l\times h)=2h(l+b)$

Length of diagonal of a cuboid =√(l${}^{}$+ b${}^{}$+ h${}^{}$)

Cube and its Surface Area

For a cube, length = breadth = height

ac

Cube with length l

TSA (cube) $=2\times \left(3l^2\right)=6l^2$
Similarly, the Lateral surface area of cube $=2(l\times l+l\times l)=4l^2$
Note: Diagonal of a cube $=\surd 3l$

Cylinder and its Surface Area

Take a cylinder of base radius r and height h units. The curved surface of this cylinder, if opened along the diameter (d = 2r) of the circular base can be transformed into a rectangle of length $2\pi r$ and height h units. Thus,

e Areas and Volumes Class 

10 Notes

SCSA of a cylinder of bas

CSA of a cylinder of base radius r and height $h=2\pi \times r\times h$
TSA  of a cylinder of base radius r and height $h=2\pi \times r\times h$ + area of two circular bases
$=2\pi \times r\times h+2\pi r^2$
$=2\pi r(h+r)$

Right Circular Cone and its Surface Area

Consider a right circular cone with slant length l, radius r and height h

Cone with base radius r and height h

CSA of right circular cone $=\pi rl$
TSA = CSA + area of base $=\pi rl+\pi r^2=\pi r(l+r)$

Sphere and its Surface Area

For a sphere of radius r

Curved Surface Area (CSA) = Total Surface Area (TSA) = $4\pi r^2$

Volume of a Cuboid

Volume of a cuboid $=\left(basearea\right)\times height=\left(lb\right)h=lbh$

Volume of a Cube

Volume of a cube = $basearea\times height$
Since all dimensions of a cube are identical, volume = $l^3$
Where l is the length of the edge of the cube.

Volume of a Cylinder

Volume of a cylinder = Base area $\times$height = $\left(\pi r^2\right)\times h=\pi r^{2}h$

Cylinder with height h and base radius r 

Volume of a Right Circular Cone

The volume of a Right circular cone is 1/3 times that of a cylinder of same height and base.
In other words, 3 cones make a cylinder of the same height and base.
The volume of a Right circular cone $=$$\mathrm{(1/3)\pi }{r}^{2}h$
Where r is the radius of the base and h is the height of the cone.

.

The volume of a Sphere

The volume of a sphere of radius r = (4/3)$\pi r^3$

Hemisphere and its Surface Area

Hemisphere of radius r

 

We know that the CSA of a sphere  $=4\pi r^2$.

A hemisphere is half of a sphere.
$\therefore$ CSA of a hemisphere of radius r $=2\pi r^2$
Total Surface Area = curved surface area + area of the base circle
$\Rightarrow$TSA $=3\pi r^2$

Volume of Hemisphere

The volume (V) of a hemisphere will be half of that of a sphere.
$\therefore$ The volume of the hemisphere of radius r $=$$\mathrm{(2/3)\pi }{r}^3$

IMPORTANT QUESTIONS FOR BOARD EXAMS

Question 1.
In Figure, a decorative block, made up of two solids a cube and a hemisphere. The base of the block is a cube of side 6 cm and the hemisphere fixed on the top has a diameter of 3.5 cm. Find the total surface area of the block.(use π=22/7)

Question 2.
A well of diameter 4 m is dug 21 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 3 m to form an embankment. Find the height of the embankment

Question 3.
The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 sq. cm, find the volume of the cylinder, (use π=22/7)

Question 4.
A right circular cone of radius 3 cm, has a curved surface area of 47.1 cm2. Find the volume of the cone, (use π = 3.14)

Question 5.
A toy is in the form of a cone of base radius 3.5 cm mounted on a hemisphere of base diameter 7 cm. If the total height of the toy is 15.5 cm, find the total surface area of the toy(use π =22/7)

Question 6.
A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3×5/9 cm. Find the diameter of the cylindrical vessel

Question 7.
A cylindrical tub, whose diameter is 12 cm and height 15 cm is full of ice-cream. The whole ice-cream is to be divided into 10 children in equal ice-cream cones, with conical base surmounted by hemispherical top. If the height of conical portion is twice the diameter of base, find the diameter of conical part of ice-cream cone.

Question 8.
Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively offered to the state government to provide place and the canvas for 1500 tents to be fixed by the government and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs ? 120 per sq.m, find the amount shared by each school to set up the tents.What value is genetated by above problem?

Question 9.
In figure, from a cuboidal solid metallic block of dimensions 15 cm X 10 cm X 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block.

Question 10.
Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m, and the canvas to be used costs 100 rupees per sq. m, find the amount, the associations will have to pay.What values are shown by these associations ?

Question 11.
504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area

Question 12.
A 21 m deep well with diameter 6 m is dug and the earth from digging is evenly spread to form a platform 27 m x 11 m. Find the height of the platform

Question 13.
If the total surface area of a solid hemisphere is 462 cm², find its volume.

Question 14.
Water in a canal, 6 m wide and 1.5 m deep, is flowing at a speed of 4 km/h. How much area will it irrigate in 10 minutes, if 8 cm of standing water is needed for irrigation?

Question 15.
A 5 m wide cloth is used to make a conical tent of base diameter 14 m and height 24 m.Find the cost of cloth used at the rate 25 rupees /metre.

Question 16

A solid sphere of radius 10.5 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed.

HOPE YOU LIKE THE QUESTIONS...................

SARASWATI TUITION CENTRE