| Exercise
14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
Exercise 14.4
Question 1
The radii of the ends of a frustum of a cone 45 cm high are 28cm and 7cm. Find its volume, the curved surface area. [Take π=22/7]Sol :
Frustum = difference of two right circular cones OAB and OCD
Let the height of the cone OAB be h1 and its slant height l1
i.e. OA = OB = l1 and OP = h1.
Let h2 be the height of cone OCD and l2 its slant height
i.e. OC = OD = l2 and OQ = h2
We have , r1 = 28cm and r2 = 7cm
and height of frustum (h) = 45cm
Also,
h1 = 45 + h2 …(i)
Now, we first need to determine the h1 and h2
∵ ΔOPB and OQD are similar, we have
$\frac{h_{1}}{h_{2}}=\frac{28}{7}=\frac{4}{1}$
⇒ h1 = 4h2 …(ii)
From (i) and (ii), we get
4h2 = 45 + h2
⇒ 3h2 = 45
⇒ h2 = 15cm
Putting the value of h2 in eq. (ii), we get
h1 = 4×15 = 60cm
So, h1 = 60cm and h2 = 15cm
Now, the volume of frustum = Vol. of cone OAB – Vol. of cone OCD
$=\frac{1}{3} \pi r_{1}^{2} h_{1}-\frac{1}{3} \pi r_{2}^{2} h_{2}$
$=\frac{1}{3} \times \frac{22}{7}\left[28^{2} \times 60-7^{2} \times 15\right]$
= 48510 cm3
Now, we first have to find the slant height l1 and l2
l1 = √{(28)2+(60)2}
⇒ l1 = √{(4×7)2 + (4×15)2
⇒ l1 = 4√{(7)2 + (15)2}
⇒ l1 = 4√49 + 225
⇒ l1 = 4√(274)
⇒ l1 = 4 × 16.55
⇒ l1 = 66.20cm
and l2 = √{(7)2 + (15)2}
⇒ l2 = √49 + 225
⇒ l2 = √(274)
⇒ l2 = 16.55 cm
So, CSA of frustum = CSA of cone OAB – CSA of cone OCD
= πr1l1 – πr2l2
$=\frac{22}{7}[28 \times 66.20-7 \times 16.55]$
= 22 [4×66.20 – 16.55]
= 22 × 248.25
= 5461.5cm2
Sol :
Given: Height of frustum = 14cm
Diameter of 1st circular end = 4cm
So, radius (R) = 2cm
Diameter of 2nd circular end = 2cm
So, radius (r) = 1cm
Capacity of the glass = Volume of frustum
Hence, the volume of glass = 102.66cm3
Greater radius = R = 33 cm
Smaller radius = r = 27 cm
Slant height = l = 10 cm
Total Surface area of the frustum = πR2 + πr2 + πl(R+r)
= π[(33)2 + (27)2+ 10×(33+27)]
= π[1089 + 729 + 600]
$=\frac{22}{7} \times 2418$
Sol :
Let radii of the circular ends of the frustum are R and r respectively.
Given that Perimeter of one end = 96cm
⇒ 2πR = 96
Perimeter of other end = 68cm
⇒ 2πr = 68
It is given that height of the frustum = 20cm
So, Slant height, l = √{h2 + (R – r)2}
= √{(20)2 +(15.27 – 10.82)2
= √400 + (4.45)2
= √400 + 19.80
= √419.80
= 20.49 cm
Now,
Volume of the frustum $=\frac{1}{3} \pi \mathrm{h}\left(\mathrm{R}^{2}+\mathrm{r}^{2}+\mathrm{Rr}\right)$
$=\frac{1}{3} \times \frac{22}{7} \times 20\left[(15.27)^{2}+(10.82)^{2}+15.27 \times 10.82\right]$
$=\frac{22 \times 20}{21}[233.1729+117.0724+165.2214]$
$=\frac{22 \times 20 \times 515.4667}{21}$
= 10800.25 cm3
Total Surface Area of the frustum = πR2 + πr2 + πl(R+r)
= π[(15.27)2 + (10.82)2+ 20.49×(15.27+10.82)]
= π[233.1729 + 117.0724 + 534.5841]
$=\frac{22}{7} \times 884.83$
= 2780.89 cm2
Sol :
We have,
Diameter of one end = 10cm
So, radius (R) = 5cm
Diameter of other end = 8cm
So, radius (r) = 4cm
and slant height, l = 8cm
So, height, h = √{l2 – (R – r)2}
= √{(8)2 – (5 – 4)2
= √64 – (1)2
= √64 – 1
= √63
= 7.937 cm
Bearing surface of the clutch = CSA of the frustum
= πl (R + r)
= 3.14 × 8 (5 + 4)
= 25.12 (9)
= 226.08 cm2
Volume of the frustum $=\frac{1}{3} \pi \mathrm{h}\left(\mathrm{R}^{2}+\mathrm{r}^{2}+\mathrm{Rr}\right)$
$=\frac{1}{3} \times 3.14 \times 7.93\left[(5)^{2}+(4)^{2}+5 \times 4\right]$
$=\frac{3.14 \times 7.93}{3}[25+16+20]$
$=\frac{3.14 \times 7.937 \times 61}{3}$
= 506.75 cm3
Hence, Volume of frustum = 506.75 cm3
[Take π=22/7]
Sol :
Greater diameter of the frustum = 56 cm
So, radius of the frustum = R = 28 cm
Smaller diameter of the frustum = 42 cm
So, Radius of the smaller end of the frustum = r = 21 cm
and Height of the frustum = h = 15 cm
$=\frac{1}{3} \pi r_{1}^{2} h_{1}-\frac{1}{3} \pi r_{2}^{2} h_{2}$
$=\frac{1}{3} \times \frac{22}{7}\left[28^{2} \times 60-7^{2} \times 15\right]$
$=\frac{1}{3} \times \frac{22}{7} \times 15\left[28^{2} \times 4-7^{2}\right]$
$=5 \times \frac{22}{7}[3136-49]$
$=5 \times \frac{22}{7} \times 3087$
= 5 × 22 × 441= 48510 cm3
Now, we first have to find the slant height l1 and l2
l1 = √{(28)2+(60)2}
⇒ l1 = √{(4×7)2 + (4×15)2
⇒ l1 = 4√{(7)2 + (15)2}
⇒ l1 = 4√49 + 225
⇒ l1 = 4√(274)
⇒ l1 = 4 × 16.55
⇒ l1 = 66.20cm
and l2 = √{(7)2 + (15)2}
⇒ l2 = √49 + 225
⇒ l2 = √(274)
⇒ l2 = 16.55 cm
So, CSA of frustum = CSA of cone OAB – CSA of cone OCD
= πr1l1 – πr2l2
$=\frac{22}{7}[28 \times 66.20-7 \times 16.55]$
= 22 [4×66.20 – 16.55]
= 22 × 248.25
= 5461.5cm2
Question 2
A drinking glass is in the shape of a frustum of a cone of height 14cm. The diameters of its two circular ends are 4cm and 2cm. Find the capacity of the glass.Sol :
Given: Height of frustum = 14cm
Diameter of 1st circular end = 4cm
So, radius (R) = 2cm
Diameter of 2nd circular end = 2cm
So, radius (r) = 1cm
Capacity of the glass = Volume of frustum
$=\frac{1}{3} \pi \mathrm{h}\left(\mathrm{R}^{2}+\mathrm{r}^{2}+\mathrm{Rr}\right)$
$=\frac{1}{3} \times \frac{22}{7} \times 14\left[(2)^{2}+(1)^{2}+2 \times 1\right]$
$=\frac{44}{3}[4+1+2]$
$=\frac{44 \times 7}{3}$
= 102.66 cm3Hence, the volume of glass = 102.66cm3
Question 3
The radii of the circular ends of a solid frustum of a cone are 33 cm and 27cm, and its slant height is
10cm. Find its capacity and total surface area. [Take π=22/7]
Sol :Greater radius = R = 33 cm
Smaller radius = r = 27 cm
Slant height = l = 10 cm
Total Surface area of the frustum = πR2 + πr2 + πl(R+r)
= π[(33)2 + (27)2+ 10×(33+27)]
= π[1089 + 729 + 600]
$=\frac{22}{7} \times 2418$
= 7599.428 cm2
= 7599.43 cm2
= 7599.43 cm2
We first need to find the height, h
l2 = (R – r)2 + h2
⇒ (10)2 = (33 – 27)2 + h2
⇒ 100 = (6)2 + h2
⇒ 100 = 36 + h2
⇒ h2 = 100 – 36
⇒ h2 = 64
⇒ h = ±8
⇒ h = 8cm
l2 = (R – r)2 + h2
⇒ (10)2 = (33 – 27)2 + h2
⇒ 100 = (6)2 + h2
⇒ 100 = 36 + h2
⇒ h2 = 100 – 36
⇒ h2 = 64
⇒ h = ±8
⇒ h = 8cm
Now, Capacity of a solid frustum of a cone = Volume of frustum
$=\frac{1}{3} \pi
\mathrm{h}\left(\mathrm{R}^{2}+\mathrm{r}^{2}+\mathrm{Rr}\right)$
$=\frac{1}{3} \times \frac{22}{7} \times
8\left[(33)^{2}+(27)^{2}+33 \times 27\right]$
$=\frac{22 \times 8}{21}[1089+729+891]$
$=\frac{22 \times 8 \times 2709}{21}$
= 22704 cm3
Question 4
The perimeters of the ends of the frustum of a cone are 96 cm and 68 cm. If the height of the frustum be 20 cm, find its radii, slant height, volume and total surface. [Take π=22/7]Sol :
Let radii of the circular ends of the frustum are R and r respectively.
Given that Perimeter of one end = 96cm
⇒ 2πR = 96
$\Rightarrow \mathrm{R}=\frac{96}{2 \pi}$
$\Rightarrow \mathrm{R}=\frac{96 \times 7}{2 \times 22}$
⇒ R = 15.27 cmPerimeter of other end = 68cm
⇒ 2πr = 68
$\Rightarrow \mathrm{r}=\frac{68}{2 \pi}$
$\Rightarrow \mathrm{r}=\frac{68 \times 7}{2 \times 22}$
⇒ r = 10.82 cmIt is given that height of the frustum = 20cm
So, Slant height, l = √{h2 + (R – r)2}
= √{(20)2 +(15.27 – 10.82)2
= √400 + (4.45)2
= √400 + 19.80
= √419.80
= 20.49 cm
Now,
Volume of the frustum $=\frac{1}{3} \pi \mathrm{h}\left(\mathrm{R}^{2}+\mathrm{r}^{2}+\mathrm{Rr}\right)$
$=\frac{1}{3} \times \frac{22}{7} \times 20\left[(15.27)^{2}+(10.82)^{2}+15.27 \times 10.82\right]$
$=\frac{22 \times 20}{21}[233.1729+117.0724+165.2214]$
$=\frac{22 \times 20 \times 515.4667}{21}$
= 10800.25 cm3
Total Surface Area of the frustum = πR2 + πr2 + πl(R+r)
= π[(15.27)2 + (10.82)2+ 20.49×(15.27+10.82)]
= π[233.1729 + 117.0724 + 534.5841]
$=\frac{22}{7} \times 884.83$
= 2780.89 cm2
Question 5
A friction clutch in the form of the frustum of a cone, the diameters of the ends being 8cm, and 10 cm and length 8 cm. Find its bearing surface and its volume. [Take π=3.14]Sol :
We have,
Diameter of one end = 10cm
So, radius (R) = 5cm
Diameter of other end = 8cm
So, radius (r) = 4cm
and slant height, l = 8cm
So, height, h = √{l2 – (R – r)2}
= √{(8)2 – (5 – 4)2
= √64 – (1)2
= √64 – 1
= √63
= 7.937 cm
Bearing surface of the clutch = CSA of the frustum
= πl (R + r)
= 3.14 × 8 (5 + 4)
= 25.12 (9)
= 226.08 cm2
Volume of the frustum $=\frac{1}{3} \pi \mathrm{h}\left(\mathrm{R}^{2}+\mathrm{r}^{2}+\mathrm{Rr}\right)$
$=\frac{1}{3} \times 3.14 \times 7.93\left[(5)^{2}+(4)^{2}+5 \times 4\right]$
$=\frac{3.14 \times 7.93}{3}[25+16+20]$
$=\frac{3.14 \times 7.937 \times 61}{3}$
= 506.75 cm3
Hence, Volume of frustum = 506.75 cm3
Question 6
A bucket is in the form of a frustum of a cone. Its depth is 15cm and the diameters of the top and the bottom are 56 cm and 42 cm respectively. Find how many litres of water can the bucket hold.[Take π=22/7]
Sol :
Greater diameter of the frustum = 56 cm
So, radius of the frustum = R = 28 cm
Smaller diameter of the frustum = 42 cm
So, Radius of the smaller end of the frustum = r = 21 cm
and Height of the frustum = h = 15 cm
Capacity of the frustum
$=$ Volume of the frustum $=\frac{1}{3} \pi \mathrm{h}\left(\mathrm{R}^{2}+\mathrm{r}^{2}+\mathrm{Rr}\right)$
$=\frac{1}{3} \times \frac{22}{7} \times 15 \times\left[28^{2}+21^{2}+28 \times 21\right]$
$=$ Volume of the frustum $=\frac{1}{3} \pi \mathrm{h}\left(\mathrm{R}^{2}+\mathrm{r}^{2}+\mathrm{Rr}\right)$
$=\frac{1}{3} \times \frac{22}{7} \times 15 \times\left[28^{2}+21^{2}+28 \times 21\right]$
$=\frac{22 \times 15}{21}[784+441+588]$
$=22 \times \frac{5}{7} \times 1813$
= 28490 cm3= 28.49 litres
Question 7
An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made
of the same metallic sheet (see Fig.) The diameters of the two circular ends of the bucket are 45 cm and
25cm, the total vertical height of the bucket is 40cm and that of the cylindrical base is 6cm. Find the area
of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket.
Also, find the volume of water the bucket can hold. [Take π=22/7]
Diameter of bigger circular end = 45cm
So, Radius $(\mathrm{R})=\frac{45}{2} \mathrm{cm}$
Diameter of smaller circular end = 25cm
So, Radius $(\mathrm{r})=\frac{25}{2} \mathrm{cm}$
Now,
Height of frustum = Total height of bucket – Height of cylinder
= 40 – 6
= 34cm
Also,
Slant height of the frustum, l = √{h2 + (R – r)2}
= √ 1256
= 35.44cm
Curved Surface Area of frustum = πl(R + r)
= 3898.4 cm2
Curved Surface area of cylinder = 2πrh
$=2 \times \frac{22}{7} \times \frac{25}{2} \times 6$
= 471.428 cm2
Area of circular base = πr2
$=\frac{22}{7} \times \frac{25}{2} \times \frac{25}{2}$
= 491.07 cm2
Now, Area of metallic sheet used
= CSA of frustum + Area of circular base + CSA of cylinder
= 3898.4 + 471.428 + 491.07
= 4860.89
= 4860.9 cm2
Volume of water that bucket can hold $=\frac{1}{3} \pi \mathrm{h}\left(\mathrm{R}^{2}+\mathrm{r}^{2}+\mathrm{Rr}\right)$
So, Radius $(\mathrm{r})=\frac{25}{2} \mathrm{cm}$
Now,
Height of frustum = Total height of bucket – Height of cylinder
= 40 – 6
= 34cm
Also,
Slant height of the frustum, l = √{h2 + (R – r)2}
$=\sqrt{34^{2}+\left(\frac{45}{2}-\frac{25}{2}\right)^{2}}$
$=\sqrt{1156+\left(\frac{20}{2}\right)^{2}}$
= √1156 + 100= √ 1256
= 35.44cm
Curved Surface Area of frustum = πl(R + r)
$=\frac{22}{7} \times 35.44 \times\left(\frac{45}{2}+\frac{25}{2}\right)$
$=\frac{22}{7} \times 35.44 \times \frac{70}{2}$
= 22 × 35.44 × 5= 3898.4 cm2
Curved Surface area of cylinder = 2πrh
$=2 \times \frac{22}{7} \times \frac{25}{2} \times 6$
= 471.428 cm2
Area of circular base = πr2
$=\frac{22}{7} \times \frac{25}{2} \times \frac{25}{2}$
= 491.07 cm2
Now, Area of metallic sheet used
= CSA of frustum + Area of circular base + CSA of cylinder
= 3898.4 + 471.428 + 491.07
= 4860.89
= 4860.9 cm2
Volume of water that bucket can hold $=\frac{1}{3} \pi \mathrm{h}\left(\mathrm{R}^{2}+\mathrm{r}^{2}+\mathrm{Rr}\right)$
$=\frac{1}{3} \times \frac{22}{7} \times 34\left[(22.5)^{2}+(12.5)^{2}+22.5 \times
12.5\right]$
$=\frac{22 \times 34}{21}[506.25+156.25+281.25]$
$=\frac{22 \times 34 \times 943.75}{21}$
= 33615.47 cm3We know that 1 cm3 = 0.001
litre
∴ Volume of water that bucket can hold = 33.62 litres (approx.)
Greater diameter of the bucket = 30 cm
Radius of the bigger end of the bucket = R = 15 cm
Diameter of the smaller end of the bucket = 10 cm
Radius of the smaller end of the bucket = r = 5 cm
Height of the bucket = 24 cm
Slant height, l = √{h2 + (R – r)2}
= √{(24)2 +(15 – 5)2
= √576 + (10)2
= √576 + 100
= √676
= 26 cm
Now,
Volume of the frustum $=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)$ $=\frac{1}{3} \times 3.14 \times 24\left[(15)^{2}+(5)^{2}+15 \times 5\right]$
= 3.14 × 8 [225 + 25 + 75]
= 3.14 × 8 ×325
= 8164 cm3
= 8.164 Litres
A litre of milk cost Rs 20
So, total cost of filling the bucket with milk = Rs 20 × 8.164
= Rs 163.28
Surface Area of the bucket
= CSA of the frustum + Area of circular base
= πl(R+r) + πr2
= π [{26×(15+5)} + (5)2]
= π[520 + 25 ]
= 3.14 × 545
= 1711.3 cm2
∴ Volume of water that bucket can hold = 33.62 litres (approx.)
Question 8
A bucket made up of a metal sheet is in the form of frustum of a cone. Its depth is 24 cm and the
diameters of the top and bottom are 30 cm and 10 cm respectively. Find the cost of milk which can completely
fill the bucket at the rate of Rs. 20 per litre and the cost of metal sheet used if it costs Rs. 10 per
100cm2.
[Use π=3.14]
Sol :Greater diameter of the bucket = 30 cm
Radius of the bigger end of the bucket = R = 15 cm
Diameter of the smaller end of the bucket = 10 cm
Radius of the smaller end of the bucket = r = 5 cm
Height of the bucket = 24 cm
Slant height, l = √{h2 + (R – r)2}
= √{(24)2 +(15 – 5)2
= √576 + (10)2
= √576 + 100
= √676
= 26 cm
Now,
Volume of the frustum $=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)$ $=\frac{1}{3} \times 3.14 \times 24\left[(15)^{2}+(5)^{2}+15 \times 5\right]$
= 3.14 × 8 [225 + 25 + 75]
= 3.14 × 8 ×325
= 8164 cm3
= 8.164 Litres
A litre of milk cost Rs 20
So, total cost of filling the bucket with milk = Rs 20 × 8.164
= Rs 163.28
Surface Area of the bucket
= CSA of the frustum + Area of circular base
= πl(R+r) + πr2
= π [{26×(15+5)} + (5)2]
= π[520 + 25 ]
= 3.14 × 545
= 1711.3 cm2
Cost of 100cm2 of metal sheet =
Rs 10
Cost of $1 \mathrm{cm}^{2}$ of metal sheet $=\frac{10}{100}$ cost of $1711.3 \mathrm{cm}^{2}$ of metal sheet $=\frac{1711.3 \times 10}{100}$
= Rs 171.13
[Take π=22/7]
Sol :
For the lower portion of the tent:
Diameter of the base = 14 m
Radius, R, of the base = 7 m
Diameter of the top end of the frustum = 7 m
Cost of $1 \mathrm{cm}^{2}$ of metal sheet $=\frac{10}{100}$ cost of $1711.3 \mathrm{cm}^{2}$ of metal sheet $=\frac{1711.3 \times 10}{100}$
= Rs 171.13
Question 9
A tent is made in the form of a conic frustum, surmounted by a cone. The diameters of the base and top of the frustum are 14m and 7 m and its height is 8 m. The height of the tent is 12m. Find the quantity of canvas required.[Take π=22/7]
Sol :
For the lower portion of the tent:
Diameter of the base = 14 m
Radius, R, of the base = 7 m
Diameter of the top end of the frustum = 7 m
Radius of the top end of the frustum =$\mathrm{r}=\frac{7}{2} \mathrm{m}=3.5 \mathrm{m}$
Height of the frustum = h = 8 m
Slant height = l = √{h2 + (R – r)2}
= √{(8)2 +(7 – 3.5)2
= √64 + (3.5)2
= √64 + 12.25
= √76.25
= 8.73 m
For the conical part
Radius of the cone base = r = 3.5 m
Height of the cone = Height of the tent – height of frustum
= 12 – 8
= 4 m
Slant height of cone = L = √(4)2+(3.5)2
= √16 + 12.25
=√28.25
= 5.3 m
Total quantity of canvas = CSA of frustum + CSA of conical top
= πl(R + r) + πrL
= π [8.73(7 + 3.5) + 3.5 × 5.3]
Diameter of top of funnel = 10cm
So, Radius (R) = 5cm
Slant height = l = √{h2 + (R – r)2}
= √{(8)2 +(7 – 3.5)2
= √64 + (3.5)2
= √64 + 12.25
= √76.25
= 8.73 m
For the conical part
Radius of the cone base = r = 3.5 m
Height of the cone = Height of the tent – height of frustum
= 12 – 8
= 4 m
Slant height of cone = L = √(4)2+(3.5)2
= √16 + 12.25
=√28.25
= 5.3 m
Total quantity of canvas = CSA of frustum + CSA of conical top
= πl(R + r) + πrL
= π [8.73(7 + 3.5) + 3.5 × 5.3]
$=\frac{22}{7}[91.66+18.6]$
$=\frac{22}{7}[110.26]$
= 346.5 m2Question 10
An oil funnel of tin sheet consists of a cylindrical portion 8 cm along attached to a frustum of a cone.
If the total height be 15cm, the diameter of the cylindrical portion 1 cm and diameter of the top of the
funnel 10cm, find the area of the tin required.
[Take π=22/7]
Sol :Diameter of top of funnel = 10cm
So, Radius (R) = 5cm
Diameter of cylindrical portion = 1cm
So, radius $(\mathrm{r})=\frac{1}{2}=0.5 \mathrm{cm}$
Height of frustum = Total height – Height of cylindrical part
= 15 – 8
= 7cm
and Slant Height, l = √{h2 + (R – r)2}
= √{(7)2 +(5 – 0.5)2
= √49 + (4.5)2
= √49 + 20.25
= √69.25
= 8.32 cm
Now, CSA of frustum = πl (R + r)
So, radius $(\mathrm{r})=\frac{1}{2}=0.5 \mathrm{cm}$
Height of frustum = Total height – Height of cylindrical part
= 15 – 8
= 7cm
and Slant Height, l = √{h2 + (R – r)2}
= √{(7)2 +(5 – 0.5)2
= √49 + (4.5)2
= √49 + 20.25
= √69.25
= 8.32 cm
Now, CSA of frustum = πl (R + r)
$=\frac{22}{7} \times 8.32 \times(5+0.5)$
$=\frac{22}{7} \times 8.32 \times(5.5)$
= 143.81 cm2Height of cylinder, H = 8cm
Now, CSA of cylinder = 2πrH
$=2 \times \frac{22}{7} \times 0.5 \times 8$
= 25.14 cm2
Area of tin required = CSA of frustum + CSA of cylinder
= 143.81 + 25.14
= 168.95 cm2
Now, CSA of cylinder = 2πrH
$=2 \times \frac{22}{7} \times 0.5 \times 8$
= 25.14 cm2
Area of tin required = CSA of frustum + CSA of cylinder
= 143.81 + 25.14
= 168.95 cm2
No comments:
Post a Comment