Surface Areas & Volume
Mathematics Class Tenth
NCERT Exercise 13.4
Take `pi = 22/7`, unless stated otherwise.
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:
Given, height of frustum of a cone = 14 cm
Diameter (bigger) = 4 cm
Radius, `r_1=4/2=2\ cm`
Diameter (smaller) = 2 cm
And Radius, `r_2 = 2/2=1\ cm`
Volume of the glass, i.e. capacity of the glass =?
Volume of a frustum of a cone `=1/3pih(r_1^2+r_2^2+r_1*r_2)`
`=1/3xx22/7xx14(2^2+1^2+2xx1) cm^3`
`=1/3xx22xx2(4+1+2)cm^3`
`=44/3xx7\ cm^3`
`=308/3\ cm^3`
`=102\2/3\ cm^3`
Thus, capacity of the glass `=102\2/3\ cm^3` Answer
Question: 2. The slant height of a frustum of a cone is 4 cm and he perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
Given, Slant height of the frustum, `l =4\ cm`
Perimeter (Circumference) of bigger end =18 cm
Perimeter (Circumference) of smaller end = 6 cm
Curved surface area (CSA) of the frustum =?
Circumference `=2pir`
For bigger circular end
`=>18\ cm = 2\ pi r_1`
`=>r_1 = 18/(2\ pi)cm`
`=r_1= 9/pi\ cm`
For smaller circular end
Circumference `=2\ pi\ r`
`=>6\ cm = 2\ pi\ r_2`
`=>r_2 = 6/(2\ pi)cm`
`=>r_2 = 3/pi\ cm`
Now, curved surface area of a frustum of cone `=pi(r_1+r_2)l`
`=pi\ (9/pi+3/pi)xx4\ cm^2`
`=pixx1/pi(9+3)xx4\ cm^2`
`=48\ cm^2`
Thus, curved surface area of the given frustum `=48\ cm^2` Answer
Question: 3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). 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:
Given, Slant height of fez (frustum), `l=15\ cm`
Radius of open side, i.e. radius of bigger end, `r_1 = 10\ cm`
Radius of smaller end, `r_2 = 4\ cm`
Material used to make the fez (frustum(Turks cap))=?
Area of a circle `=pir^2`
Thus, area of upper base of fez `=pi (4\ cm)^2`
`=16\ pi\ cm^2`
Curved surface area of frustum`=pi(r_1+r_2)l`
`=pi(10\ cm+4\ cm)xx15\ cm`
`=pi\ 14xx15\ cm^2`
`=210\ pi\ cm^2`
Thus, Material required to make the cap = total surface area of cap
= curved surface area of cap + area of upper base
`=210\ pi\ cm^2 + 16\ pi\ cm^2`
`=pi(210+16)cm^2`
`=22/7xx226\ cm^2`
`=4972/7\ cm^2`
`=710\2/7\ cm^2`
Thus, material required to make the given fez `=710\2/7\ cm^2` Answer
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 `cm^2`. (Take `pi = 3.14`)
Solution:
Given, height of the frustum =16 cm
Upper radius of container `r_1=20\ cm`
Lower radius of container, `r_2=8\ cm`
Cost of milk = Rs 20 per litre
Cost of metal sheet = Rs 8 per `100\ cm^2`
Cost of milk to fill the container =?
Cost of metal sheet =?
Here to find the cost of milk, the volume of container (in the shape of frustum) is to be calculated. And to find the cost of metal sheet, the total surface area of the container is needed to be calculated.
Volume of a frustum of a cone `=1/3pih(r_1^2+r_2^2+r_1*r_2)`
Thus, volume of given container in the shape of frustum
`=1/3xx22/7xx16(20^2+8^2+(20xx8))cm^3`
`=352/21(400+64+160)cm^3`
`=(352xx624)21\ cm^3`
`=219648/21 cm^3`
`=10459.43\ cm^3`
∵ cost of 1 litre or `1000\ cm^3` of milk =Rs 20
∴ cost of `1\ cm^3` of milk `=20/1000` Rs
∴ cost of `10459.43\ cm^3` of milk = Rs `(20xx10459.43)/1000`
=Rs 209.18
Now, Area of circle `=pir^2`
Thus, area of lower end of container `=pi xx 8^2\ cm^2`
`=64\ pi\ cm^2`
Slant height of a frustum of a cone`l=sqrt(h^2+(r_1-r_2)^2)`
`=>l = sqrt(16^2+(20-8)^2)cm`
`=>l=sqrt(256+12^2)cm`
`=>l=sqrt(256+144)cm`
`=>l=sqrt(400)cm`
`=>l=20\ cm`
Curved surface area of frustum`=pi(r_1+r_2)l`
Thus, curved surface area of given frustum `=pi(20+8)xx20\ cm^2`
`=pixx28xx20\ cm^2`
`=pi\ 560\ cm^2`
Thus, total surface area of given frustum = curved surface area + area of base
`= pi\ 560\ cm^2+64\ pi\ cm^2`
`=pi(560+64) cm^2`
`=3.14 xx 624\ cm^2`
`=1959.36\ cm^2`
Now, ∵ cost of `100\ cm^2` of metal sheet = Rs 8
∴ cost of `1\ m^2` of metal sheet `=8/100` Rs
∴ cost of `1959.36\ cm^2` of metal sheet `=8/100 xx 1959.36` Rs
= Rs 156.75
Thus, cost of milk is Rs 209 and cost of metal sheet is Rs 156.75 Answer
Question: 5. A metallic right circular cone 20 cm high and whose vertical angle is `60^0` 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:
Given, Height of cone `=20\ cm`
Vertical angle `/_ BAC =60^0`
Given cone is cut into two parts at the middle of its height and a frustum is made.
Thus, length of wire having diameter of `1/16\ cm` =?
Now, since, cone is cut into two parts from middle of its height, thus, height of frustum `=10\ cm`
Thus, in figure,
`AF=FG = 10\ cm`
`AG = 20\ cm`
`/_BAC =60^0`
Thus, `/_DAF = 30^0`
In `triangle` DAF
`tan\ 30^0 = (DF)/(AF)`
`=>1/sqrt3 = (DF)/10`
`=>DF = 10/sqrt3`
`=>DF = r_2=10/sqrt3`
Now, in `triangle` BAG
`tan\ 30^0 = (BG)/(AG)`
`=>1/sqrt3 = (BG)/20`
`=>BG =20/sqrt3`
`=>BG = R_1 =20/sqrt3`
Now, For resulting frustum,
Height = 10 cm
Bigger Radius `r_1=20/sqrt3`
And smaller Radius, `r_2 = 10/sqrt3`
Volume of a frustum of a cone `=1/3pih(r_1^2+r_2^2+r_1*r_2)`
Thus, volume of the resulting frustum
`=1/3\ pixx 10((20/sqrt3)^2+(10/sqrt3)^2+20/sqrt3 xx 10/sqrt3)cm^3`
`=1/3 pixx10(400/3+100/3+200/3)cm^3`
`=1/3\ pixx10((400+100+200)/3)cm^3`
`=1/3xx22/7xx10xx700/3 cm^3`
`=1/3xx22xx10xx100/3 cm^3`
`=22000/9 cm^3`
Or, Volume of resulting frustum `=22000/9 cm^3`
Volume of cylinder `=pi r^2h`
Since, wire forms a shape of cylinder
Let, length of wire `=l`
Given, diameter of wire `=1/16\ cm`
∴ radius of wire `=1/16xx1/2=1/32 cm`
Thus, volume of wire `=22/7 xx (1/32)^2xxl`
`=22/7xx1/32xx1/32xxl cm^2`
`=22/(7168)l\ cm^2`
Since wire is to be drawn from resulting frustum
Thus, volume of wire = volume of frustum
`=>22/(7168)l\ cm^2=22000/9 cm^3`
`=>l = (22000\ cm^3)/9 xx (7168)/(22\ cm^2)`
`=>l = (1000xx7168)/9 cm`
`=>l =796444.44\ cm`
`=>l = (796444.44)/100\ m`
`=>l = 7964.44\ m`
Thus, length of the wire = 7964.44 meter. Answer
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