Chapter 7 – Mensuration
Practice set 7.1
1. Find the volume of a cone if the radius of its base is 1.5 cm and its perpendicular height is 5 cm.
Solution:
For the cone.
radius (r) = 1.5 cm, height (h) = 5 cm
Volume (V)
= \(\large \frac {1}{3}\)πr²h
= \(\large \frac {1}{3}\) × 3.14 × 1.5 × 1.5 × 5
= 11.775 cm³
Ans: Volume of the cone is 11.775 cm³.
2. Find the volume of a sphere of diameter 6 cm.
Solution:
For the sphere,
Diameter = 6 cm
∴ Radius (r) = \(\large \frac {6}{2}\) cm = 3 cm
Volume of the sphere
= \(\large \frac {4}{3}\)πr³
= \(\large \frac {4}{3}\) × 3.14 × 3 × 3 × 3
= 113.04 cm³
Ans: Volume of the sphere 113. 04 cm³.
3. Find the total surface area of a cylinder if the radius of its base is 5 cm and height is 40 cm.
Solution:
For a cylinder,
r = 5 cm …[Given]
h = 40 cm …[Given]
Total surface area
= 2πr (r + h)
= 2 × 3.14 × 5 × (5 + 40)
= 31.4 × 45
= 1413 cm²
Ans: Total surface area of the cylinder is 1,413 cm².
4. Find the surface area of a sphere of radius 7 cm.
Solution:
For the sphere,
Radius (r) = 7 cm
Curved surface area of the sphere
= 4πr²
= 4 × \(\large \frac {22}{7}\) × 7 × 7
= 616 cm²
Ans: Curved surface area of the sphere is 616 cm².
5. The dimensions of a cuboid are 44 cm, 21 cm, 12 cm. It is melted and a cone of height 24 cm is made. Find the radius of its base.
Solution:
For the solid cuboid,
l = 44 cm, b = 21 cm, h = 12 cm
For the solid cone.
h₁ = 24 cm
r = ?
Cone is made by melting the cuboid.
∴ Volume of cone = volume of cuboid.
\(\large \frac {1}{3}\)πr²h₁ = l × b × h
\(\large \frac {1}{3}\) × \(\large \frac {22}{7}\) × r² × 24 = 44 × 21 × 12
∴ r² = \(\large \frac {44\,×\,22\,×\,12\,×\,3\,×\,7}{22\,×\,24}\)
∴ r² = 21 × 3 × 7
∴ r² = 21 × 21
∴ r = 21 cm …[Taking square roots]
Ans: Radius of the cone is 21 cm.
6. Observe the measures of pots in figure 7.8 and 7.9. How many jugs of water can the cylindrical pot hold ?
Solution:
For the conical watering,
r = 3.5 cm, h = 10 cm
For the cylindrical water pot,
r₁ = 7 cm, h₁ = 10 cm
Let ‘N’ be the number of jugs required to fill the cylindrical pot completely.
∴ N × Volume of cone = Volume of cylinder
∴ N × \(\large \frac {1}{3}\)πr²h = πr₁²h
∴ N × \(\large \frac {1}{3}\)π × 3.5 × 3.5 × 10 = π × 7 × 7 × 10
∴ N = \(\large \frac {π\,×\,7\,×\,7\,×\,10\,×\,3}{π\,×\,3.5\,×\,3.5\,×\,10}\)
∴ N = 12
Ans: Number of conical jugs required to fill up the cylindrical pot completely is 12.
7. A cylinder and a cone have equal bases. The height of the cylinder is 3 cm and the area of its base is 100 cm².The cone is placed upon the cylinder. Volume of the solid figure formed is 500 cm³. Find the total height of the figure.
Solution:
Let the radius of base of each part be r.
Height of the cylinder (h₁) = 3 cm.
Let the height of the cone be h₂
Area of the base = 100 cm²
∴ πr²h = 100 cm² …(i)
Volume of the solid figure formed = Volume of the cylinder + Volume of the cone
∴ 500 = πr²h₁ + \(\large \frac {1}{3}\)πr²h₂
∴ 500 = πr² \(\large (\)h₁ + \(\large \frac {1}{3}\)h₂\(\large )\) …[From (i)]
∴ 500 = 100 \(\large (\)3 + \(\large \frac {h₂}{3})\)
∴ \(\large \frac {500}{100}\) = 3 + \(\large \frac {h₂}{3}\)
∴ 5 = 3 + \(\large \frac {h₂}{3}\)
∴ 5 – 3 = \(\large \frac {h₂}{3}\)
∴ \(\large \frac {h₂}{3}\) = 2
∴ h₂ = 6
Total height = h₁ + h₂ = 3 + 6 = 9 cm
Ans: Total height of the figure is 9 cm.
8. In figure 7.11, a toy made from a hemisphere, a cylinder and a cone is shown. Find the total area of the toy.
Solution:
Toy is made up of a cone, cylinder and hemisphere of equal radii.
For the conical part
Radius (r) = 3 cm
Height (h) = 4 cm
For the cylindrical part
Radius (r) = 3 cm
height (h₁) = 40 cm
For hemisphere
Radius (r) = 3cm
Let the slant height of the conical part be l.
∴ l² = r² + h²
∴ l² = 3² + 4²
∴ l² = 9 + 16
∴ l² = 25
∴ l = 5 cm (Taking square roots)
Total surface area of the toy
= Curved surface area of the hemisphere + Curved surface area of the cylinder + Curved surface area of the cone
= 2πr² + 2πrh₁ + πrl
= πr(2r + 2h₁ + l)
= \(\large \frac {22}{7}\) × 3 × (2 × 3 + 2 × 40 + 5)
= \(\large \frac {22}{7}\) × 3 × 91
= 22 × 3 × 13
= 858 cm²
Ans: Total Surface area of the toy is 858 sq.cm
9. In the figure 7.12, a cylindrical a wrapper of flat tablets is shown. The radius of a tablet is 7 mm and its thickness is 5 mm. How many such tablets are wrapped in the wrapper?
Solution:
For cylindrical wrapper,
Diameter = 14 mm
Radius (R) = \(\large \frac {14}{2}\) mm = 7 mm
Height (H) = 10 cm = 100 mm
For cylindrical tablet,
Radius (r) = 7 mm, Height (h) = 5 mm
Let ‘N’ number of tablets can be wrapped in the given wrapper.
∴ N × Volume of tablet = Volume of wrapper.
∴ N × πr²h = πr²H
∴ N × π × 7 × 7 × 5 = π × 7 × 7 × 100
∴ N = \(\large \frac {π\,×\,7\,×\,7\,×\,100}{π\,×\,7\,×\,7\,×\,5}\)
∴ N = 20
Ans: 20 tablets can be packed in the given wrapper.
10. Figure 7.13 shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and the surface area of the toy from the measures shown in the figure. (π = 3.14)
Solution:
For the conical part
Radius (r) = 3 cm
Height (h) = 4 cm
Let l be the slant height of conical part
l² = r² + h²
∴ l² = 3² + 4²
∴ l² = 9 + 16
∴ l² = 25
∴ l = 5 cm …[Taking square root]
For the hemisphere
Radius (r) = 3 cm
Volume of the toy
= Volume of the Cone + volume of the hemisphere
= \(\large \frac {1}{3}\)πr²h + \(\large \frac {2}{3}\)πr³
= \(\large \frac {1}{3}\)πr² (h + 2r)
= \(\large \frac {1}{3}\) × 3.14 × 3 × 3 (4 + 2 × 3)
= 3.14 × 3 (4 + 6)
= 3.14 × 3 × 10
= 3.14 × 30
= 94.2 cm³
∴ Volume of the toy is 94.2cm³
Surface area of toy
= Curved surface area of the cone + Curved surface area of the hemisphere
= πrl + 2πr²
= πr(l + 2r)
= 3.14 × 3 (5 + 2 × 3)
= 3.14 × 3 (5 + 6)
= 3.14 × 3 × 11
= 3.14 × 33
= 103.62 cm².
Ans: Surface area of the toy is 103.62 cm².
11. Find the surface area and the volume of a beach ball shown in the figure.
Solution:
For a spherical beach ball.
Diameter = 42 cm …[Given]
Radius (r) = \(\large \frac {42}{2}\) = 21 cm
Volume of the spherical beach ball
= \(\large \frac {4}{3}\)πr³
= \(\large \frac {4}{3}\) × \(\large \frac {22}{7}\) × 21 × 21 × 21
= 38,808 cm³
Surface area of the spherical beach ball
= 4πr²
= 4 × \(\large \frac {22}{7}\) × 21 × 21
= 5,544 cm²
Ans: Volume of the spherical beach ball is 38,808 cm³ and surface area of the spherical beach ball is 5,544 cm².
12. As shown in the figure, a cylindrical glass contains water. A metal sphere of diameter 2 cm is immersed in it. Find the volume of the water.
Solution:
For the cylindrical vessel
Diameter = 14 cm
Radius (r₁) = 7cm
Height of the water level (h₁) = 30 cm
When sphere is immersed:
For the Sphere
Diameter = 2 cm
Radius (r₂) = 1 cm
Apparent volume of water when sphere is immersed in water (V₁)
= πr₁²h₁
= \(\large \frac {22}{7}\) × 7 × 7 × 30
= 4620 cm³
Volume of the Sphere (V₂)
= \(\large \frac {4}{3}\)πr₂³
= \(\large \frac {4}{3}\) × 3.14 × 1 × 1 × 1
= 4.19 cm³
Actual volume of water
= V₁ – V₂
= 4620 – 4.19
= 4615.81 cm³
Ans: Actual volume of water is 4615.81 cm³
Practice set 7.2
1. The radii of two circular ends of a frustum shaped bucket are 14 cm and 7 cm. Height of the bucket is 30 cm. How many liters of water can it hold ? (1 litre = 1000 cm³)
Solution:
For a frustum shaped bucket,
radius of bigger circle (r₁) = 14 cm
radius of smaller circle (r₂) = 7 cm
height (h) = 30 cm
Capacity of the bucket
= Volume of the bucket
= \(\large \frac {1}{3}\)πh (r₁² + r₂² + r₁ × r₂)
= \(\large \frac {1}{3}\) × \(\large \frac {22}{7}\) × 30 (14² + 7² + 14 × 7)
= \(\large \frac {22}{7}\) × 10 (186 + 49 + 98)
= \(\large \frac {22}{7}\) × 10 × 343
= 22 × 10 × 49
= 10,780 cm³
= 10780 × 1000 …[∵ 1 litre = 1000 cm³]
= 10.78 litres
Ans: Capacity of the bucket is 10.780 litres.
2. The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its
i) curved surface area
ii) total surface area.
iii) volume (π = 3.14)
Solution:
For a frustum
radius of bigger circle (r₁) = 14 cm
radius of smaller circle (r₂) = 6 cm
height (h) = 6 cm
Slant height of the frustum (l)
= \(\sqrt{h²\,+\,(r₁\,–\,r₂)²}\)
= \(\sqrt{6²\,+\,(14\,–\,6)²}\)
= \(\sqrt{6²\,+\,8²}\)
= \(\sqrt{36\,+\,64}\)
= \(\sqrt{100}\)
= 10 cm …[Taking square root]
(i) Curved surface area of the frustum
= π (r₁ + r₂) l
= 3.14 × (14 + 6) × 10
= 3.14 × 20 × 10
= 628 cm²
Ans: Curved surface area of the frustum is 628 cm²
(ii) Total surface area of the frustum
= π (r₁ + r₂)l + πr₁² + πr₂²
= 628 + π (14² + 6²)
= 628 + π (196 + 36)
= 628 + 3.14 × 232
= 628 + 728.48
= 1,356.48 cm²
Ans: Total surface area of the frustum is 1356.48 cm².
(iii) Volume of the frustum
= \(\large \frac {1}{3}\)πh (r₁² + r₂² + r₁ × r₂)
= \(\large \frac {1}{3}\) × 3.14 × 6 (14² + 6² + 14 × 6)
= 3.14 × 2 (196 + 36 + 84)
= 3.14 × 2 × 316
= 1,984.48 cm³
Ans: Volume of the frustum is 1,984.48 cm³.
3. The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of the frustum complete the following activity. (π = \(\large \frac {22}{7}\)).
circumference₁ = 2πr₁ = 132
∴ r₁ = \(\large \frac {132}{2π}\) = □
circumference₂ = 2πr₂ = 132
∴ r₂ = \(\large \frac {132}{2π}\) = □
slant height of frustum, l
= \(\sqrt{h²\,+\,(r₁\,–\,r₂)²}\)
= \(\sqrt{□²\,+\,□²}\)
= □ cm
curved surface area of the frustum
= π(r₁+ r₂)l
= π × □ × □
= □ sq.cm.
Solution:
circumference₁ = 2πr₁ = 132
∴ r₁ = \(\large \frac {132}{2π}\) = 21 cm
circumference₂ = 2πr₂ = 132
∴ r₂ = \(\large \frac {132}{2π}\) = 14 cm
slant height of frustum, l = \(\sqrt{h²\,+\,(r₁\,–\,r₂)²}\)
l = \(\sqrt{24²\,+\,(21\,–\,14)²}\)
l = \(\sqrt{24²\,+\,7²}\)
l = \(\sqrt{576\,+\,49}\)
l = \(\sqrt{625}\)
l = 25 cm
Curved surface area of the frustum
= π(r₁ + r₂)l
= π × 35 × 25
= 2750 sq.cm.
Practice set 7.3
1. Radius of a circle is 10 cm. Measure of an arc of the circle is 54°. Find the area of the sector associated with the arc. (π = 3.14)
Solution:
For the sector,
r = 10 cm, θ = 54°
Area of the sector
= \(\large \frac {θ}{360}\) × πr²
= \(\large \frac {54}{360}\) × 3.14 × 10 × 10
= \(\large \frac {3}{20}\) × 314
= \(\large \frac {942}{20}\)
= 47.1 cm²
Ans: Area of the sector is 47.1 cm².
2. Measure of an arc of a circle is 80 cm and its radius is 18 cm. Find the length of the arc. (π = 3.14)
Solution:
For a sector,
r = 18 cm, θ = 80
Length of an arc
= \(\large \frac {θ}{360}\) × 2πr
= \(\large \frac {80}{360}\) × 2 × 3.14 × 17
= 3.14 × 8
= 25.12 cm
Ans: Length of the arc is 25.12 cm.
3. Radius of a sector of a circle is 3.5 cm and length of its arc is 2.2 cm. Find the area of the sector.
Solution:
For the sector,
r = 3.5 cm, length of arc (l) = 2.2 cm
Area of the sector
= l × \(\large \frac {r}{2}\)
= 2.2 × \(\large \frac {3.5}{2}\)
= 3.85 cm²
Ans: Area of the sector is 3.85 cm².
4. Radius of a circle is 10 cm. Area of a sector of the circle is 100 cm². Find the area of its corresponding major sector. (π = 3.14)
Solution:
For the circle, r = 10 cm
Area of minor sector = 100 cm²
Area of the circle
= πr²
= 3.14 × 10 × 10
= 314 cm²
Area of a major circle
= Area of the circle – Area of corresponding minor sector
= 314 – 100
= 214 cm²
Ans: Area of the major sector is = 214 cm².
5. Area of a sector of a circle of radius 15 cm is 30 cm². Find the length of the arc of the sector.
Solution:
For the circle, r = 15 cm
Area of the sector = 30 cm²
Area of the sector = Length of arc × \(\large \frac {r}{2}\)
∴ 30 = Length of arc × \(\large \frac {15}{2}\)
∴ Length of arc = \(\large \frac {30\,×\,2}{15}\)
∴ Length of arc = 4 cm
Ans: Length of the arc is 4 cm
6. In the figure 7.31, radius of the circle is 7 cm and m(arc MBN) = 60°, find
(1) Area of the circle.
(2) A(O – MBN).
(3) A(O – MCN).
Solution:
For the circle, r = 7 cm
m(arc MBN) = θ = 60°
(i) Area of the circle
= πr²
= \(\large \frac {22}{7}\) × 7 × 7
= 154 cm²
Ans: Area of the circle is 154 cm²
(ii) A (sector O – MBN)
= \(\large \frac {θ}{360}\) × πr²
= \(\large \frac {60}{360}\) × 154
= \(\large \frac {1}{6}\) × 154
= 25.67 cm²
Ans: A (sector O – MBN) is 25.67 cm²
(iii) A (sector O – MCN)
= Area of the circle – A (sector O – MBN) = 154 – 25.67
= 128.33 cm²
Ans: A (sector O – MCN) is 128.33 cm²
7. In figure 7.32, the radius of the circle is 3.4 cm and the perimeter of sector P – ABC is 12.8 cm. Find A(P – ABC).
Solution:
For the circle, r = 3.4 cm
Perimeter of sector P – ABC = 12.8 cm
∴ P(P – ABC) = Length of arc (l) + r + r
∴ 12.8 = l + 3.4 + 3.4
∴ 12.8 – 6.8 = l
∴ l = 6 cm
Area of the sector
= l × \(\large \frac {r}{2}\)
= 6 × \(\large \frac {3.4}{2}\)
= 10.2 cm²
Ans: Area of the sector is 10.2 cm²
8. In figure 7.33 O is the centre of the sector. ∠ROQ = ∠ MON = 60°. OR = 7 cm, and OM = 21 cm. Find the lengths of arc RXQ and arc MYN. (π = \(\large \frac {22}{7}\))
Solution:
(i) For arc RXQ,
θ = ∠ROQ = 60°
OR (r) = 7 cm
Length of arc RXQ
= \(\large \frac {θ}{360}\) × 2πr
= \(\large \frac {60}{360}\) × 2 × \(\large \frac {22}{7}\) × 7
= 7.33 cm
Ans: Length of arc RXQ is 7.33 cm
(ii) For arc MYN,
OM (r) = 21 cm,
θ = ∠MON = 60°
Length of arc MYN
= \(\large \frac {θ}{360}\) × 2πr
= \(\large \frac {60}{360}\) × 2 × \(\large \frac {22}{7}\) × 21
= 22 cm
Ans: Length of arc (MYN) is 22 cm
9. In figure 7.34, if A(P-ABC) = 154 cm² radius of the circle is 14 cm, find
(1) ∠APC.
(2) l(arc ABC).
Solution:
Region P – ABC is a sector
A (P – ABC) = \(\large \frac {θ}{360}\) × πr²
∴ 154 = \(\large \frac {θ}{360}\) × \(\large \frac {22}{7}\) × 14 × 14
∴ \(\large \frac {154\,×\,360\,×\,7}{22\,×\,14\,×\,14}\) = θ
∴ θ = 90°
∴ ∠APC = 90°
Length of arc ABC
= \(\large \frac {θ}{360}\) × 2πr
= \(\large \frac {90}{360}\) × 2 × \(\large \frac {22}{7}\) × 14
= 22 cm
Ans: Length of arc ABC is 22 cm.
10. Radius of a sector of a circle is 7 cm. If measure of arc of the sector is –
(1) 30°
(2) 210°
(3) three right angles;
find the area of the sector in each case.
Solution:
For the circle, r = 7 cm
(i) For the sector,
θ = 30°
Area of sector
= \(\large \frac {θ}{360}\) × πr²
= \(\large \frac {30}{360}\) × \(\large \frac {22}{7}\) × 7 × 7
= 12.83 cm²
Ans: Area of the sector is 12.83 cm².
(ii) For the sector,
θ = 210°
Area of sector
= \(\large \frac {θ}{360}\) × πr²
= \(\large \frac {210}{360}\) × \(\large \frac {22}{7}\) × 7 × 7
= 89.83
Ans: Area of sector is 89.83 cm².
(iii) For the sector,
θ = 3 right angles = 3 × 90° = 270°
Area of sector
= \(\large \frac {θ}{360}\) × πr²
= \(\large \frac {270}{360}\) × \(\large \frac {22}{7}\) × 7 × 7
= 115.5 cm²
Ans: Area of the sector is 115.50 cm².
11. The area of a minor sector of a circle is 3.85 cm² and the measure of its central angle is 36°. Find the radius of the circle.
Solution:
For the sector,
Area = 3.85 cm² , θ = 36°
Area of minor sector = \(\large \frac {θ}{360}\) × πr²
∴ 3.85 = \(\large \frac {36}{360}\) × \(\large \frac {22}{7}\) × r²
∴ \(\large \frac {3.85\,×\,360\,×\,7}{36\,×\,22}\) = r²
∴ r² = 12.25
∴ r = 3.5 cm
Ans: Radius of the circle is 3.5 cm.
12. In figure 7.35, □ PQRS is a rectangle. If PQ = 14 cm, QR = 21 cm, find the areas of the parts x, y and z.
Solution:
□PQRS is a rectangle with l = 21cm and b = 14cm
A (□PQRS)
= l × b
= 21 × 14
A (□PQRS) = 294 cm² …(i)
For region x ie. for sector Q – PT,
r = 14 cm, θ = 90°
A (region x)
= A (sector Q – PT)
= \(\large \frac {θ}{360}\) × πr²
= \(\large \frac {90}{360}\) × \(\large \frac {22}{7}\) × 14 × 14
= 154 cm² …(ii)
Ans: A (region x) is 154 cm²
QP = QT …[Radii of the same circle]
∴ QT = 14 cm …(iii)
QR = QT + RT …[Q – T – R]
21 = 14 + RT …[From (iii) and given]
RT = 21 – 14 = 7 cm …(iv)
For region y i.e. for (sector R – B T)
r₁ = 7cm, θ = 90°
A (region y)
= A (sector R – B T)
= \(\large \frac {θ}{360}\) × πr²
= \(\large \frac {90}{360}\) × \(\large \frac {22}{7}\) × 14 × 14
= 38.5 cm²
Ans: A (region y) is 38.5 cm².
A (rectangle PQRS) = A (region x) + A (region y) + A (region z) …[Area addition property]
294 = 154 + A (region z) + 38.5
A (region z) = 294 – 154 – 38.5 …[From (i), (ii) and (iv)]
A (region z) = 101.5 cm²
Ans: A (region z) is 101.5 cm²
13. ∆LMN is an equilateral triangle. LM = 14 cm. As shown in figure, three sectors are drawn with vertices as centres and radius 7 cm.
Find,
(1) A (∆ LMN)
(2) Area of any one of the sectors.
(3) Total area of all the three sectors.
(4) Area of the shaded region.
Solution:
(i) ∆LMN is an equilateral triangle with side 14 cm
A (∆LMN)
= \(\large \frac {\sqrt{3}}{4}\) × side²
= \(\large \frac {1.73}{4}\) × 14 × 14
= 84.77 cm² …(i)
Ans: A (∆LMN) is 84.77 cm²
(ii) For sector L – AB,
r = 7 cm, θ = 60° …[Angle of an equilateral triangle]
A (sector L – AB)
= \(\large \frac {θ}{360}\) × πr²
= \(\large \frac {60}{360}\) × \(\large \frac {22}{7}\) × 7 × 7
= \(\large \frac {77}{3}\)
= 25.67 cm²
Ans: Area of a sector is 25.67 cm².
(iii) For all three sectors, radii and central angles are equal
∴ Area of all sectors are equal.
∴ Total of areas of all three sectors
= 3 × A (sector A – LB)
= 3 × \(\large \frac {77}{3}\)
= 77 cm²
Ans: Total areas of all three sectors = 77 cm²
(iv) Area of the shaded region
= A (∆LMN) – Area of three sectors
= 84.77 – 77 …[From (i) and (ii)]
= 7.77 cm²
Ans: Area of the shaded region is 7.77 cm².
Practice set 7.4
1. In figure 7.43, A is the centre of the circle. ∠ABC = 45° and AC = 7 2 cm. Find the area of segment BXC.
Solution:
In ∆ABC, AB = AC …[Radii of same circle]
∴ ∠ABC = ∠ACB …[Isosceles triangle theorem]
But, ∠ABC = 45° …[Given]
∴ ∠ACB = 45° …(i)
In ∆ABC,
∠ACB = ∠ABC = 45° …[From (i) and Given]
∴ ∠BAC = 90° …[Remaining angle of ∆ABC]
For segment BXC,
θ = 90°, r = 7 2 cm
Arc of segment BXC)
= r² \(\large [\frac {πθ}{360}\) – \(\large \frac {sinθ}{2}]\)
= \(7\sqrt{2}\)² \(\large [\frac {3.14\,×\,90}{360}\) – \(\large \frac {sin\,90}{2}]\)
= 98 × \(\large [\frac {1.57}{2}\) – \(\large \frac {1}{2}]\)
= 98 × \(\large [\frac {1.57\,–\,1}{2}]\)
= 98 × \(\large [\frac {0.57}{2}]\)
= 49 × 0.57
= 27.93
Ans: Arc of segment BXC is 27.93 sq cm.
2. In the figure 7.44, O is the centre of the circle. m(arc PQR) = 60°, OP = 10 cm. Find the area of the shaded region. (π = 3.14, \(\sqrt{3}\) = 1.73)
Solution:
m (arc PQR) = m ∠POR …[Definition of measure of minor arc]
m∠PQR = 60° …(i)
For segment PQR,
r = OP = 10 cm
θ = m ∠POR = 60° …[From (i)]
Area of shaded portion
= A (segment PQR)
= r² \(\large [\frac {πθ}{360}\) – \(\large \frac {sinθ}{2}]\)
= (10)² \(\large [\frac {3.14\,×\,60}{360}\) – \(\large \frac {sin\,60}{2}]\)
= 100 × \(\large [\frac {3.14}{6}\) – \(\large \frac {\sqrt{3}}{2\,×\,2}]\)
= 100 × \(\large [\frac {3.14\,×\,2}{6\,×\,2}\) – \(\large \frac {1.73\,×\,3}{4\,×\,3}]\)
= 100 × \(\large [\frac {6.28\,–\,5.19}{12}]\)
= 98 × \(\large \frac {1.09}{12}\)
= 9.08 cm²
Ans: Area of shaded portion is 9.08 cm².
3. In the figure 7.45, if A is the centre of the circle. ∠PAR = 30°, AP = 7.5, find the area of the segment PQR. (π = 3.14)
Solution:
For segment PQR,
r = AP = 7.5 units
θ = ∠PAR = 30°
A (segment PQR)
= r² \(\large [\frac {πθ}{360}\) – \(\large \frac {sinθ}{2}]\)
= (7.5)² \(\large [\frac {3.14\,×\,30}{360}\) – \(\large \frac {sin\,30}{2}]\)
= 56.25 × \(\large [\frac {3.14}{12}\) – \(\large \frac {1}{2\,×\,2}]\)
= 56.25 × \(\large [\frac {3.14\,–\,3}{12}]\)
= 56.25 × \(\large \frac {0.14}{12}\)
= 0.66 cm²
Ans: A (segment PQR) is 0.66 sq. units.
4. In the figure 7.46, if O is the centre of the circle, PQ is a chord. ∠POQ = 90°, area of shaded region is 114 cm², find the radius of the circle. (π = 3.14)
Solution:
For segment PRQ,
θ = ∠POQ = 90°
A (segment PRQ) = 114 cm²
A (segment PQR) = r² \(\large [\frac {πθ}{360}\) – \(\large \frac {sinθ}{2}]\)
∴ 114 = r² \(\large [\frac {3.14\,×\,90}{360}\) – \(\large \frac {sin\,90}{2}]\)
∴ 114 = r² × \(\large [\frac {3.14}{4}\) – \(\large \frac {1}{2}]\)
∴ 114 = r² × \(\large [\frac {1.57\,–\,1}{2}]\)
∴ 114 = r² × \(\large \frac {0.57}{2}\)
∴ \(\large \frac {114\,×\,2}{0.57}\) = r²
∴ r² = \(\large \frac {114\,×\,2\,×\,100}{57}\)
∴ r² = 2 × 2 × 10 × 10
∴ r = 20 …[Taking square roots]
Ans: Radius of the circle is 20 cm.
5. A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centre of the circle. Find the area of the minor as well as the major segment. (π = 3.14, \(\sqrt{3}\) = 1.73)
Solution:
For minor segment,
r = 15 cm and θ = 60°
Area of minor segment
= r² \(\large [\frac {πθ}{360}\) – \(\large \frac {sinθ}{2}]\)
= (15)² \(\large [\frac {3.14\,×\,60}{360}\) – \(\large \frac {sin\,60}{2}]\)
= 225 × \(\large [\frac {3.14}{6}\) – \(\large \frac {\sqrt{3}}{2\,×\,2}]\)
= 225 × \(\large [\frac {3.14\,×\,2}{6\,×\,2}\) – \(\large \frac {1.73\,×\,3}{4\,×\,3}]\)
= 225 × \(\large [\frac {6.28\,–\,5.19}{12}]\)
= 225 × \(\large \frac {1.09}{12}\)
= 20.44 cm²
Ans: Area of minor segment is 20.44 cm².
Area of circle
= πr²
= 3.14 × 15 × 15
= 706.5 cm²
Ans: Area of the circle is 706.5 cm².
Area of major segment
= Area of circle – Area of minor segment
= 706.5 – 20.44
= 686.06 cm²
Ans: Area of the major segment is 686.06 cm².
Problem Set 7
1. Choose the correct alternative answer for each of the following questions.
(1) The ratio of circumference and area of a circle is 2:7. Find its circumference.
(A) 14π
(B) \(\large \frac {7}{π}\)
(C) 7π
(D) \(\large \frac {14}{π}\)
Ans: Option (A) : 14π
(2) If measure of an arc of a circle is 160° and its length is 44 cm, find the circumference of the circle.
(A) 66 cm
(B) 44 cm
(C) 160 cm
(D) 99 cm
Ans: Option (D) : 99 cm
(3) Find the perimeter of a sector of a circle if its measure is 90 ° and radius is 7 cm.
(A) 44 cm
(B) 25 cm
(C) 36 cm
(D) 56 cm
Ans: Option (B) : 25 cm
(4) Find the curved surface area of a cone of radius 7 cm and height 24 cm.
(A) 440 cm²
(B) 550 cm²
(C) 330 cm²
(D) 110 cm²
Ans: Option (B) : 550 cm²
(5) The curved surface area of a cylinder is 440 cm² and its radius is 5 cm. Find its height.
(A) 44 cm
(B) 22 cm
(C) 44 cm
(D) 22 cm
Ans: Option (A) : 44 cm
(6) A cone was melted and cast into a cylinder of the same radius as that of the base of the cone. If the height of the cylinder is 5 cm, find the height of the cone.
(A) 15 cm
(B) 10 cm
(C) 18 cm
(D) 5 cm
Ans: Option (A) : 15 cm
(7) Find the volume of a cube of side 0.01 cm.
(A) 1 cm³
(B) 0.001 cm³
(C) 0.0001 cm³
(D) 0.000001 cm³
Ans: Option (D) : 0.000001 cm³
(8) Find the side of a cube of volume 1 m³.
(A) 1 cm
(B) 10 cm
(C) 100 cm
(D) 1000 cm
Ans: Option (C) : 100 cm
2. A washing tub in the shape of a frustum of a cone has a height of 21 cm. The radii of the circular top and bottom are 20 cm and 15 cm respectively. What is the capacity of the tub? (π = \(\large \frac {22}{7})\)
Solution:
For frustum shaped tub,
r₁ = 20 cm, r₂ = 15 cm, h = 21 cm
Quantity of water that can be contained in the tub
= Inner volume of tub
= \(\large \frac {1}{3}\) × \(\large \frac {22}{7}\) × h (r₁² + r₂² + r₁ × r₂)
= \(\large \frac {1}{3}\) × \(\large \frac {22}{7}\) × 21 (20² + 15² + 20 × 15)
= 22 (400 + 225 + 300)
= 22 × 925
= 20,350 cm³
= \(\large \frac {20350}{1000}\) litres …[∵1 litres = 1000 cm]
= 20.35 litres
Ans: Quantity of water that can be contained in the tub is 20.35 litres.
3. Some plastic balls of radius 1 cm were melted and cast into a tube. The thickness, length and outer radius of the tube were 2 cm, 90 cm and 30 cm respectively. How many balls were melted to make the tube?
Solution:
For spherical solid ball, r = 1 cm,
For cylindrical pipe, Outer radius (r₁) = 30 cm
Thickness (t) = 2 cm
Height (h) = 90 cm
Inner radius (r₂)
= r₁ – t
= 30 – 2
∴ r₂ = 28 cm
Volume of cylindrical pipe = Number of spherical balls required (N) × Volume of spherical ball
π × h(r₁² – r₂²) = N × \(\large \frac {4}{3}\) πr³
∴ π × 90(30² – 28²) = N × \(\large \frac {4}{3}\) π(1)³
∴ π × 90(900 – 784) = N × \(\large \frac {4}{3}\) π
∴ π × 90(116) = N × \(\large \frac {4}{3}\) π
∴ N = \(\large \frac {π\,×\,90\,×\,116\,×\,3}{4\,×\,π\,×\,2}\)
∴ N = 7,830
Ans: Number of spherical balls required is 7,830.
4. A metal parallelopiped of measures 16 cm × 11cm × 10 cm was melted to make coins. How many coins were made if the thickness and diameter of each coin was 2 mm and 2 cm respectively?
Solution:
For the metallic cuboid,
l = 16 cm, b = 11 cm, h = 10 cm
For the cylindrical coin,
Diameter = 2 cm, Thickness (h₁) = 2 mm = 0.2 cm
i.e. Radius (r₁) = 1 cm
Let number of coins made be N.
∴ N × Volume of coin = Volume of cuboid
∴ N × πr₁² = l × b × h
∴ N × \(\large \frac {22}{7}\) × 1 × 1 × \(\large \frac {2}{10}\) = 16 × 11 × 10
∴ N = \(\large \frac {16\,×\,11\,×\,10\,×\,10\,×\,7}{22\,×\,2}\)
∴ N = 2800
Ans: Number of coins made is 2800.
5. The diameter and length of a roller is 120 cm and 84 cm respectively. To level the ground, 200 rotations of the roller are required. Find the expenditure to level the ground at the rate of Rs. 10 per sq.m.
Solution:
For circular roller,
Diameter = 120 cm,
∴ radius (r) = \(\large \frac {120}{2}\) = 60 cm
length (h) = 84 cm
Number of rotations required to level the ground (N) = 200
Rate of levelling (R) = ₹10 per sq.m
Area levelled in 1 rotation = curved surface area of the roller.
∴ Area levelled in 200 rotations (A)
= 200 × 2πrh
= 200 × 2 × \(\large \frac {22}{7}\) × 60 × 84
= 6336000 cm²
= \(\large \frac {6336000}{100\,×\,100}\)8m²
= 633.6 m²
Cost of levelling
= A × R
= 633.6 × 10
= ₹ 6336
Ans: Cost of levelling the ground is ₹ 6336.
6. The diameter and thickness of a hollow metals sphere are 12 cm and 0.01 m respectively. The density of the metal is 8.88 gm per cm³. Find the outer surface area and mass of the sphere.
Solution:
For the metallic hollow sphere;
Outer diameter = 12 cm
∴ Outer radius (r₁) = \(\large \frac {12}{2}\) = 6 cm
Thickness
= 0.01 m
= 0.01 × 100 …[∵ 1 m = 100cm]
= 1 cm
Inner radius (r₂)
= r₂ – thickness
= 6 – 1
= 5 cm
Outer surface area of the hollow sphere
= 4πr₁²
= 4 × 3.14 × 6 × 6
= 452.16 cm²
Volume of metal in the hollow metallic sphere
= Volume of the outer sphere – Volume of the inner sphere
= \(\large \frac {4}{3}\)πr₁³ – \(\large \frac {4}{3}\)πr₂³
= \(\large \frac {4}{3}\)π (r₁³ – r₂³)
= \(\large \frac {4}{3}\) × \(\large \frac {22}{7}\) (6³ – 5³)
= \(\large \frac {4}{3}\) × \(\large \frac {22}{7}\) (216 – 125)
= \(\large \frac {4}{3}\) × \(\large \frac {22}{7}\) × 91
= \(\large \frac {4}{3}\) × 22 × 13
= \(\large \frac {1144}{3}\) cm³
Density = \(\large \frac {Mass}{Volume}\) …[Formula]
Mass = Density × Volume
Mass of the hollow sphere
= Volume of the hollow sphere × Density of Sphere
= \(\large \frac {1144}{3}\) × 8.88
= 1144 × 2.96
= 3386.24 gm
Ans: Outer surface area is 452.16 sq. cm. and mass of the hollow sphere is 3386.24 gm.
7. A cylindrical bucket of diameter 28 cm and height 20 cm was full of sand. When the sand in the bucket was poured on the ground, the sand got converted into the shape of a cone. If the height of the cone was 14 cm, what was the base area of the cone?
Solution:
For the cylindrical bucket;
diameter = 28 cm
Radius (r) = 14 cm.
height (h) = 20 cm
For conical shape sand;
height (h₁) = 14 cm
Let the radius be r₁
Sand from the bucket is emptied to form a cone.
∴ Volume of sand in the conical shape = Volume of the sand in the bucket
∴ \(\large \frac {1}{3}\)πr₁²h₁ = πr²h
∴ \(\large \frac {1}{3}\)πr₁² × 14 = \(\large \frac {22}{7}\) × 14 × 14 × 20
∴ πr₁² = \(\large \frac {22}{7}\) × 14 × 14 × 20 × \(\large \frac {14}{3}\)
∴ πr₁² = 2,640 sq.cm
Ans: Area of base of the cone is 2640 cm²
8. The radius of a metallic sphere is 9 cm. It was melted to make a wire of diameter 4 mm. Find the length of the wire.
Solution:
For the sphere;
r = 9 cm
For the wire,
Thickness (diameter) = 4 mm = 0.4 cm …[∵ 1 cm = 10 mm]
∴ Radius (r₁) = \(\large \frac {0.4}{2}\) = 2 mm = 0.2 cm
Let the length of wire be h₁
Wire is made by melting the sphere,
Volume of the wire = Volume of the sphere
πr₁² h₁ = \(\large \frac {4}{3}\)πr³
π × 0.2 × 0.2 × h₁ = \(\large \frac {4}{3}\) π × 9 × 9 × 9
h₁ = \(\large \frac {4\,×\,π\,×\,9\,×\,9\,×\,9\,×\,10\,×\,10}{3\,×\,π\,×\,2\,×\,2}\)
h₁ = 24,300 cm
h₁ = 243 m …[∵ 1 m = 100 cm]
Ans: Length of the wire formed is 243 m.
9. The area of a sector of a circle of 6 cm radius is 15π sq.cm. Find the measure of the arc and length of the arc corresponding to the sector.
Solution:
For the sector,
Radius (r) = 6 cm
Area of the sector = 15π cm²
Area of the sector = \(\large \frac {θ}{360}\) × πr²
∴ 15π = \(\large \frac {θ}{360}\) × π × 6 × 6
∴ θ = \(\large \frac {15π\,×\,360}{6\,×\,6\,×\,π}\)
∴ θ = 150
Area of the sector = l × \(\large \frac {r}{2}\)
∴ 15π = l × \(\large \frac {6}{2}\)
∴ l = \(\large \frac {15π\,×\,2}{6}\)
∴ l = 5π
Ans: Measure of the arc is 150° and length of the corresponding arc is 5π cm.
10. In the figure 7.47, seg AB is a chord of a circle with centre P. If PA = 8 cm and distance of chord AB from the centre P is 4 cm, find the area of the shaded portion. (π = 3.14, \(\sqrt{3}\) = 1.73)
Solution:
Radius of the circle is = 8 cm …(i) [Given]
i.e. PA = 8 cm
seg PM ⊥ chord AB, A – M – B …(ii)
Distance of the chord AB from the centre P is 4 cm
i.e. PM = 4 cm …(iii)
In ∆PMA,
∠PMA = 90° …[From (ii)]
∴ PM = \(\large \frac {1}{2}\) PA …[From (i) and (iii)]
∴ ∠PAM = 30° …(iv) [Converse of 30° – 60° – 90° theorem]
In ∆PMA,
∠PMA = 90° …[From (ii)]
∠PAM = 30° …[From (iv)]
∠APM = 60° …(v) [Remaining angle of triangle]
Similarly, we can prove,
∠BPM = 60° …(vi)
∠APB = ∠APM + ∠BPM …[Angle addition property]
∴ ∠APB = 60° + 60° …[From (v) and (vi)]
∴ ∠APB = 120° …(vii)
For sector (P – AXB)
θ = ∠APB = 120°
r = 8 cm
Area (sector P – AXB) = \(\large \frac {θ}{360}\) × πr²
∴ Area (sector P – AXB) = \(\large \frac {120}{360}\) × 3.14 × 8 × 8
∴ Area (sector P – AXB) = 66.99 sq cm …(viii)
∆PMA is 30° – 60° – 90° triangle …[From (ii), (iv) and (v)]
AM = \(\large \frac {\sqrt{3}}{2}\) × PA …[Side opposite to 30°]
∴ AM = \(\large \frac {\sqrt{3}}{2}\) × 8
∴ AM = 4 \(\sqrt{3}\) cm …(ix)
seg PM ⊥ chord AB .. [From (i)]
∴ AB = 2AM …[Perpendicular drawn from the centre to the chord bisects the chord]
∴ AB = 2 × 4\(\sqrt{3}\) cm …[From (ix)]
∴ AB = 8 \(\sqrt{3}\) cm …(x)
A (∆PAB)
= \(\large \frac {1}{2}\) × base × height
= \(\large \frac {1}{2}\) × AB × PM
= \(\large \frac {1}{2}\) × 8\(\sqrt{3}\) × 4
= 16 \(\sqrt{3}\)
= 16 × 1.73
= 27.68 sq cm …(xi)
A (sector P – AXB) = A (∆PAB) + A (segment AXB) …[Area addition property]
∴ 66.99 = 27.68 + A (segment AXB) …[From (viii) and (xi)]
∴ A (segment AXB) = 66.99 – 27.68
∴ A (segment AXB) = 39.28 sq. cm
Area of shaded portion = A (seg PXB) = 39.28 sq. cm
Ans: Area of the shaded portion is 39.28 sq cm.
11. In the figure 7.48, square ABCD is inscribed in the sector A-PCQ. The radius of sector C – BXD is 20 cm. Complete the following activity to find the area of shaded region.
Solution:
Side of square ABCD = radius of sector C – BXD = □ cm
Area of square = (side)² = □² = □ …(I)
Area of shaded region inside the square
= Area of square ABCD – Area of sector C – BXD
= □ – \(\large \frac {θ}{360}\) × πr²
= □ – \(\large \frac {90}{360}\) × 3.14 × 400
= □ – 314
= □
Radius of bigger sector
= Length of diagonal of square ABCD
= 20 \(\sqrt{2}\)
Area of the shaded regions outside the square
= Area of sector (A – PCQ) – Area of square ABCD
= A(A- PCQ) – A(□ABCD)
= \(\large (\frac {θ}{360}\) × πr²) – □²
= \(\large (\frac {90}{360}\) × 3.14 × (20\(\sqrt{2}\)²) – (20)²
= □ – □
= □
∴ Total area of the shaded region = 86 + 228 = 314 sq.cm.
Solution:
Side of square ABCD = radius of sector C – BXD = 20 cm
Area of square = (side)² = 20² = 400 cm² …(I)
Area of shaded region inside the square
= Area of square ABCD – Area of sector C – BXD
= 400 – \(\large \frac {θ}{360}\) × πr²
= 400 – \(\large \frac {90}{360}\) × 3.14 × 400
= 400 – 314
= 86 cm²
Radius of bigger sector
= Length of diagonal of square ABCD
= 20 \(\sqrt{2}\)
Area of the shaded regions outside the square
= Area of sector (A – PCQ) – Area of square ABCD
= A(A – PCQ) – A(□ABCD)
= \(\large (\frac {θ}{360}\) × πr²) – (side)²
= \(\large (\frac {90}{360}\) × 3.14 × (20\(\sqrt{2}\)²) – (20)²
= 628 – 400
= 228
∴ Total area of the shaded region = 86 + 228 = 314 sq.cm.
12. In the figure 7.49, two circles with centres O and P are touching internally at point A. If BQ = 9, DE = 5, complete the following activity to find the radii of the circles.
Solution:
Let the radius of the bigger circle be R and that of smaller circle be r.
OA, OB, OC and OD are the radii of the bigger circle
∴ OA = OB = OC = OD = R
PQ = PA = r
OQ = OB – BQ = □
OE = OD – DE = □
As the chords QA and EF of the circle with centre P intersect in the interior of the circle, so by the property of internal division of two chords of a circle,
OQ × OA = OE × OF
□ × R = □ × □ …(∵ OE = OF)
R² – 9R = R² – 10R + 25
R = □
AQ = 2r = AB – BQ
2r = 50 – 9 = 41
r = □ = □
Solution:
Let the radius of the bigger circle be R and that of smaller circle be r.
OA, OB, OC and OD are the radii of the bigger circle
∴ OA = OB = OC = OD = R
PQ = PA = r
OQ = OB – BQ = R – 9
OE = OD – DE = R – 5
As the chords QA and EF of the circle with centre P intersect in the interior of the circle, so by the property of internal division of two chords of a circle,
OQ × OA = OE × OF
(R – 9) × R = (R – 5) × (R – 5) …(∵ OE = OF)
R² – 9R = R² – 10R + 25
R = 25 units
AQ = 2r = AB – BQ
2r = 50 – 9 = 41
r = \(\large \frac {41}{2}\) = 20.5