Class 10 Mathematics

Surface Area Volume

Exercise 13.4 (NCERT)

Question 1: A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Solution: We have R = 2, r = 1 cm and h = 14 cm

Volume of frustum
`=1/3\πh(R^2+r^2+Rr)`

`=1/3\ πxx14(2^2+1^2+2)`

`=1/3xx22/7xx14xx7=102(2)/(3)` cm3


Question 2: The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Solution: Slant height l = 4 cm, perimeters = 18 cm and 6 cm

Radii can be calculated as follows:

`text(Radius)=(text(Preimeter))/(2π)`

Or, `R=(18)/(2π)=9/ π`

And, `r=(6)/(2 π)=3/ π`

Curved surface area of frustum

`= π(R+r)l`

`= π((9)/( π)+(3)/( π))=48` cm3


Question 3: A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution: R = 10 cm, r = 4 cm, slant height = 15 cm

Curved surface area of frustum

`= π(R+r)l`
`= π(10+4)15`
`=(22)/(7)xx14xx15=660` cm2

Area of upper base

`= πr^2= πxx4^2`
`=16 π=50(2)/(7)`

Hence, total surface area

`=660+50(2)/(7)=710(2)/(7)` cm2

Question 4: A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2.

Solution: Height of frustum = 16 cm, R = 20 cm, r = 8 cm

Volume of frustum

`1/3\ πh(R^2+r^2+Rr)`

`=1/3\ πxx16(20^2+8^2+160)`

`=1/3xx22/7xx16(400+64+160)`

`=1/3xx22/7xx16xx604=10449.92` cm2

Cost of milk @ Rs. 20 per 1000 cubic cm

`10.44992 xx 20 = Rs. 208.99`

For calculating surface area, we need to find slant height which can be calculated as follows:

`l=sqrt(h^2+(R-r)^2)`

`=sqrt(16^2+(20-8)^2)`

`=sqrt(256+144)=sqrt400=20` cm

Surface area of frustum

`= π(R+r)l+ πr^2`
`= π[(20+8)20+8^2]`

`= π(560+64)=22/7xx624=1959.36` cm2

Cost of metal sheet @ Rs. 8 per 100 sq cm `= 19.5936 xx 8 = Rs. 156.75`


Question 5: A metallic right circular cone 20 cm high and whose vertical angle is 600 is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.

Solution: Volume of frustum will be equal to the volume of wire and by using this relation we can calculate the length of the wire.

10 surface area volume frustum exercise solution

In the given figure; AO = 20 cm and hence height of frustum LO = 10 cm

In triangle AOC we have angle CAO = 300 (halft of vertical angle of cone BAC)

Therefore;

`text(tan)30°=(OC)/(AO)`

Or, `1/sqrt3=(OC)/(20)`

Or, `OC=(20)/(sqrt3)`

Using similarity cirteria in triangles AOC and ALM it can be shown that `LM = (10)/(sqrt3)` (because LM bisects the cone through its height)

Similarly, LO = 10 cm

Volume of frustum can be calculated as follows:

`V=1/3\πh(r_1^2+r_2^2+r_1\r_2)`

`=1/3\πxx10[((20)/(sqrt3))^2+((10)/(sqrt3))^2+(20)/(sqrt3)xx(10)/(sqrt3)]`

`=1/3\πxx10((400)/(3)+(100)/(3)+(200)/(3))`

`=(7000)/(9)\ π`  cm3

Volume of cylinder is given as follows:

`=πr^2h`

Or, `π(1/32)^2xxh=(7000)/(9)\ π`

Or, `h=(7000)/(9)xx1024`

`=796444.44\ cm=7964.4\ m`


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Exercise 13.1

Exercise 13.2

Exercise 13.3

Exercise 13.5