Solutions-Class-10-Mathematics-2-Chapter-7-Mensuration-Maharashtra Board

The radius (r) of the cone = 1.5 cm, Its perpendicular height (h) = 5 cm

Volume of the cone = \(\frac{1}{3}\)πr²h = \(\frac{1}{3}×\frac{22}{7}\) × 1.5 × 1.5 × 5 = \(\frac{82.5}{7}\) = 11.785 ≈ 11.79 cm³

Answer is : Volume of the cone is 11.79 cm³.

Diameter of a sphere = 6 cm, its radius(r) = 3 cm

Volume of the sphere = \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\) × 3.14 × 3 × 3 × 3 = 113.04 cm³

Answer is :  Volume of the sphere is 113.04 cm³.

The radius (r) of the base of the cylinder = 5 cm, its height (h) = 40 cm

Total surface area of the cylinder = 2πr(r + h)

= 2 × 3.14 × 5(5 + 40)

= 3.14 ×10 × 45

= 1413 cm²

Answer is : Total surface area of the cylinder is 1413 cm².

The radius of the sphere (r) = 7 cm.

Surface area of the sphere = 4πr² = 4 × \(\frac{22}{7}\) × 7 × 7 = 88 × 7 = 616 cm²

Answer is : Surface area of the sphere is 616 cm².

The length (l) of a cuboid = 44 cm, its breadth (b) = 21 cm, its height (h) = 12 cm

The height of the cone (h₁) = 24 cm

Let the radius of the base of the cone be r.

Cuboid is melted and a cone is made.

∴ volume of cuboid = volume of cone

∴ l x b x h = \(\frac{1}{3}\) x πr²h₁

∴ 44 x 21 x 12 = \(\frac{1}{3}×\frac{22}{7}\) x r² x 24

∴ r² = \(\frac{44×21×12×3×7}{22×24}\) = 21 x 21

∴ r = 21 cm     ... (Taking square roots of both the sides)

Answer is : The radius of the base of the cone is 21 cm.

The radius (r) of the conical water jug = 3.5 cm, its height (h) = 10 cm,

The radius (r₁) of the cylindrical water pot = 7 cm, its height (h₁) = 10 cm.

Number of jugs of water that cylindrical pot can hold = \(\frac{\text{Volume of cylindrical water pot}}{\text{Volume of cylindrical water jug}}\)

\(\frac{πr_1^2h_1}{\frac{1}{3}×πr^2h}\) = \(\frac{7×7×10}{\frac{1}{3}×3.5×3.5×10}\) = 2 × 2 × 3 = 12

Answer is : The cylindrical pot can hold 12 jugs of water.

A cylinder and a cone have equal bases. ∴ they have equal radii.

Let the radius be r.

Area of its base = 100 cm²

∴ πr² = 100 cm²    ….(1)

Let the height of the cylinder be h₁ and that of the cone be h₂.

h₁ = 3 cm

Volume of the solid figure = 500 cm³

∴ volume of the cylindrical part + volume of conical part = 500

∴ πr²h₁ + \(\frac{1}{3}\)πr²h₂ = 500

∴ 100 × 3 + \(\frac{1}{3}\) × 100 × h₂ = 500    ….[From (1)]

∴ 300 + \(\frac{100}{3}\)h₂ = 500

∴ \(\frac{100}{3}\)h₂ = 500 – 300 = 200

∴ h₂ = \(\frac{200×3}{100}\) = 6

∴ h₂ = 6 cm

Height of the figure = h₁ + h₂ = 3 + 6 = 9 cm.

Answer is : The total height of the figure is 9 cm.

[Note : The answer in the textbook is given as 5 cm, whereas the correct answer is 9 cm.]

A toy is made up of hemisphere, cylinder and a cone.

They have equal bases.

Let their radius be r.

∴ r = 3 cm.

Let the heights of the conical part and cylindrical part be h₁ and h₂ respectively.

h₁ = 4 cm and h₂ = 40 cm.

Let the slant height of the cone be l.

l ² = r² + h₁²

∴ l ² = 3² + 4²

∴ l ² = 9 + 16

∴ l ² = 25

∴ l = 5 cm   ... (Taking square roots of both the sides)

Total area of the toy = curved surface area of the cone + curved surface area of the cylinder + curved surface area of the hemisphere

= πrl + 2πrh₂ + 2πr²

= πr(l + 2h₂ + 2r)

= π × 3(5 + 2×40 + 2×3)

= π × 3(5 + 80 + 6)

= π × 3 (91)

= 273π cm²

Answer is : Total area of the toy is 273π cm²

The radius (r) of the tablet = 7 mm, its thickness (h) = 5mm

The diameter of the wrapper = 14 mm, ∴ its radius (r₁) = 14/2 = 7 mm

its length (h₁) = 10 cm = 100 mm

Number of tablets that can be wrapped in the wrapper = \(\frac{\text{Volume of the wrapper}}{\text{Volume of each tablet}}\)

= \(\frac{πr_1^2h_1}{πr^2h}\) = \(\frac{7×7×100}{7×7×5}\) = 20

Answer is : 20 tablets can be wrapped in the wrapper.

A toy is made up of a cone and a hemisphere.

They have equal bases.

Let the radius of the base be r.

∴ r = 3 cm

Height of the conical part (h) = 4 cm

Let the slant height of the cone be l

l² = r² + h²

∴ l² = 3² + 4²

∴ l² = 9 + 16

∴ l² = 25

∴ l = 5 ... (Taking square roots of both the sides)

Surface area of the toy = Curved surface area of the conical part + Curved surface area of the hemispherical part

= πrl + 2πr²

= πr (l + 2r)

= 3.14 × 3(5 + 2 × 3)

= 3.14 × 3 × 11

= 3.14 × 33 = 103.62 cm²

Volume of the toy = Volume of conical part + Volume of hemispherical part

= \(\frac{1}{3}\)πr²h + \(\frac{2}{3}\)πr³ = \(\frac{1}{3}\)πr²(h + 2r) = \(\frac{1}{3}\) × 3.14 × 9(4 + 2 × 3) = 94.2  cm³

Answer is : The surface area of the toy is 103.62 cm² and the volume of the toy is 94.2 cm³.

The diameter of the beach ball = 42 cm, ∴ its radius (r) = 42/2 = 21 cm

Surface area of the beach ball = 4πr² = 4 × 3.14 × 21 × 21 = 5538.96 cm²

Volume of the beach ball = \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\) × 3.14 × (21)³ = 38772.72 cm³

Answer is : Surface area of the beach ball is 5538.96 cm² and volume of the beach ball is 38772.72 cm³.

Diameter of the cylindrical glass = 14 cm, ∴ r = 7 cm

Volume of the water with metallic sphere immersed in it = πr²h = π × 49 × 30

= 1470π cm³  …..(1)

Diameter of the metallic sphere = 2 cm. ∴ r₁ = 1 cm

Volume of the metallic sphere = \(\frac{4}{3}\)πr₁³ = \(\frac{4}{3}\) × π × 1³ = \(\frac{4}{3}\)π = 1.33 π cm³  …..(2)

To find the volume of water in the cylinder, subtract the result (2) from the result (1).

∴ Volume of water = (1470 π − 1.33 π) cm³ = 1468.67 π cm³

Answer is :  Volume of water is 1468.67 π cm³.

The radii of two circular ends of frustum shape bucket are 14 cm and 7 cm.

∴ r₁ = 14 cm and r₂ = 7 cm, its height (h) = 30 cm

Volume of the bucket = \(\frac{1}{3}\)π (r₁² + r₂² + r₁ x r₂) x h

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × (142 + 72 + 14 × 7) × 30

= \(\frac{22}{7}\) × 343 × 10

= 10780 cm³ =  litres         ……(∵ 1 litre = 1000 cm³)

= 10.78 litres.

Answer is : Bucket can hold 10.78 litres of water.

The radii of ends of a frustum are 14 cm and 6 cm.

∴ r₁ = 14 cm and r₂ = 6 cm

its height (h) = 6 cm

Let the slant height of the frustum be l.

l = \(\sqrt{h^2+(r_1-r_2)^2}\)

∴ l = \(\sqrt{6^2+(14-6)^2}\)

∴ l = \(\sqrt{100}\) = 10 cm

Curved surface area of a frustum = πl(r₁ + r₂)

= 3.14 × 10 × (14 + 6) = 3.14 × 10 × 20 = 628 cm²

Total surface area of a frustum = πl(r₁ + r₂) + πr₁² + πr₂²

= 628 + π(r₁² + r₂²)

= 628 + 3.14 (14² + 6²)

= 628 + 3.14(196 + 36)

= 628 +3.14 × 232

= 628 + 728.48

= 1356.48 cm²

Volume of the frustum = \(\frac{1}{3}\)πh(r₁² + r₂² + r₁ x r₂)

= \(\frac{1}{3}\) x 3.14 × 6 ×(14² + 6² + 14 × 6)

= 3.14 × 2 × (196 + 36 + 84)

= 6.28 × 316

= 1984.48 cm³

Answer is : Curved surface area of a frustum is 628 cm², Total surface area of a frustum is 1356.48 cm² and Volume of frustum is 1984.48 cm³.

circumference₁ = 2πr₁ = 132

r₁ = \(\frac{132}{2π}\) = [21 cm]

circumference₂ = 2πr₂ = 88

r₂ = \(\frac{88}{2π}\) = [14 cm]

slant height of frustum, l = \(\sqrt{h^2+(r_1-r_2)^2}\) = \(\sqrt{[24]^2+([7])^2}\) = [25] cm

curved surface area of the frustum = π(r₁ + r₂)l

= π × [35] × [25]

= [2750] sq.cm.

The radius (r) of a circle =10 cm,

Measure of an arc of the circle (θ) = 54°

Area of the sector = \(\frac{θ}{360}\) × πr² = \(\frac{54}{360}\)  × 3.14 × 10² = 3 × 1.57 × 10 = 47.10 cm²

Answer is : Area of the sector associated with the given arc is 47.10 cm².

Measure of an arc of a circle (θ) = 80°, its radius (r) = 18 cm.

Length of the arc = \(\frac{θ}{360}\)  × 2πr = \(\frac{80}{360}\)  × 2 × 3.14 × 18 = 3.14 × 8 = 25.12 cm

Answer is : Length of the arc is 25.12 cm.

The radius (r) of the sector of a circle = 3.5 cm.

Length of the arc = 2.2 cm.

Area of a sector = \(\frac{\text{length of the arc × radius}}{2}\) = \(\frac{2.2×3.5}{2}\) = 3.85 cm²

Answer is : Area of the sector is 3.85 cm².

The radius (r) of a circle =10 cm.

Area of the circle = πr² = 3.14 × 10 × 10 = 314 cm²

Area of the sector = 100 cm²    ... (Given)

Area of corresponding major sector = Area of the circle − Area of the minor sector

= 314 – 100 = 214 cm²

Answer is : Area of the corresponding major sector is 214 cm².

The radius (r) of the circle = 15 cm.

Area of the sector = 30 cm².

Area of the sector = \(\frac{\text{length of the arc × radius}}{2}\)

30 = \(\frac{\text{length of the arc × 15}}{2}\)

∴ length of the arc = \(\frac{30×2}{15}\) = 4

∴ length of the arc = 4 cm

Answer is : Length of the arc is 4 cm.

The radius (r) of the circle = 7 cm, m(arc MBN) = θ = 60°

(1) Area of the circle = πr² = \(\frac{22}{7}\) × 7 × 7 = 154 cm²    ….(1)

(2) A(O−MBN) = \(\frac{θ}{360}\) × πr² = \(\frac{60}{360}\) × 154      …[from (1)]

                                        = \(\frac{154}{6}\)  = 25.7 cm²

(3) A(O−MCN) = Area of the circle − A(O−MBN)

= 154 − 25.7 = 128.3 cm²

Answer is : (1) Area of the circle = 154 cm², (2) A(O−MBN) = 25.7 cm² and (3) A(O−MCN) = 128.3 cm².

The radius (r) of the circle = 3.4 cm.

Perimeter of sector P−ABC = 12.8 cm.

Perimeter of sector P−ABC = r + r + length of arc ABC

∴ 12.8 =3.4 +3.4 + length of arc ABC

∴ 12.8 = 6.8 + length of arc ABC

∴ length of arc ABC = 12.8 − 6.8

∴ length of arc ABC = 6 cm

A(P−ABC) = \(\frac{\text{length of the arc × radius}}{2}\) = \(\frac{6×3.4}{2}\) = 10.2 cm²

Answer is : A(P−ABC) = 10.2 cm².

For arc RXQ,

radius(r) = OR = 7 cm

m(arc RXQ) = ∠ ROQ = θ = 60°

For arc MYN,

radius (r₁) = OM = 21 cm

m(arc MYN) = ∠ MON = θ = 60°

Length of arc RXQ = \(\frac{θ}{360}\) × 2πr = \(\frac{60}{360}\) × 2 × \(\frac{22}{7}\) × 7 = \(\frac{44}{6}\) = 7.3 cm

Length of arc MYN = \(\frac{θ}{360}\) × 2πr₁ = \(\frac{60}{360}\) × 2 × \(\frac{22}{7}\) × 21 = 22 cm

Answer is : Length of arc RXQ is 7.3 cm and length of arc MYN is 22 cm.

The radius (r) of the circle = 14 cm.

A(P−ABC) = 154 cm².

Let m(arc ABC) = APC = θ

A(P−ABC) = \(\frac{θ}{360}\) × πr²

∴ 154 = \(\frac{θ}{360}\) × \(\frac{22}{7}\) × (14)²

∴ θ = \(\frac{154×360×7}{22×14×14}\) = 90°   ∴ ∠ APC = 90°

Area of the sector = \(\frac{\text{length of the arc × radius}}{2}\)

∴ A(P−ABC) = \(\frac{\text{l(arc ABC) × r}}{2}\)

∴ 154 = \(\frac{\text{l(arc ABC) × 14}}{2}\)

∴ l(arc ABC) = \(\frac{154}{7}\) = 22 cm

Answer is : ∠ APC = 90° and l(arc ABC) = 22 cm.

(i) The radius (r) of a sector of a circle =7 cm.

Measure of an arc of the sector (θ) = 30°

Area of the sector = \(\frac{θ}{360}\) × πr² = \(\frac{30}{360}\) × \(\frac{22}{7}\) × (7)² = 12.83 cm²

(ii) The radius (r) of a sector of a circle = 7 cm.

Measure of an arc of the sector (θ) = 210°

Area of the sector = \(\frac{θ}{360}\) × πr² = \(\frac{210}{360}\) × \(\frac{22}{7}\) × (7)² =   = 89.83 cm²

(iii) The radius (r) of a sector of a circle = 7 cm.

Measure of an arc of the sector (θ) = 3 right angles = 3 × 90° = 270°

Area of the sector = \(\frac{θ}{360}\) × πr² = \(\frac{270}{360}\) × \(\frac{22}{7}\) × (7)² = \(\frac{3×22×7}{4}\) = 115.5 cm²

Answer is :

(1) Area of the sector is 12.83 cm², when measure of the arc is 30°.

(2) Area of the sector is 89.83 cm², when measure of the arc is 210°.

(3) Area of the sector is 115.5 cm², when measure of the arc is equal to three right angles.

Area of a minor sector of a circle is 3.85 cm²

Measure of its central angle (θ) = 36°.

Let its radius be r.

Area of the sector = \(\frac{θ}{360}\) × πr²

∴ 3.85 = \(\frac{36}{360}\) × \(\frac{22}{7}\) × r²

∴ r² = \(\frac{38.5×7}{22}\) = 12.25

∴ r = 3.5 cm

Answer is : The radius of the circle is 3.5 cm.

□ PQRS is a rectangle.

PQ = 14 cm and QR = 21 cm     ... (Given)

Consider points M and N as shown.

Parts x and y are sectors.

For part x,

radius (r) = PQ = 14 cm

m(arc PT) = ∠ PQT = θ = 90°

Area of part x = \(\frac{θ}{360}\) × πr² = \(\frac{90}{360}\) × \(\frac{22}{7}\) × 14² =  154 cm²

QT = PQ = 14 cm       ... (Radii of the same arc)

QT + TR = QR        ... (Q−T−R)

∴ 14 + TR = 21

∴ TR = 21 − 14

∴ TR = 7 cm

For part y,

radius (r₁) = TR = 7 cm

m(arc TU) = ∠ TRU = θ₁ = 90°

Area of part y = \(\frac{θ_1}{360}\) × πr₁² = \(\frac{90}{360}\) × \(\frac{22}{7}\) × 7² =  38.5 cm²

Area of rectangle = length x breadth

A(□ PQRS) = QR x PQ

∴ A(□ PQRS) = 21 x 14

∴ A(□ PQRS) = 294 cm²

Area of part z,

= A(□ PQRS) − Area of part x − Area of part y

= 294 – 154 − 38.5

= 101.5 cm²

Answer is : Area of part x is 154 cm², area of part y is 38.5 cm2 and area of part z is 101.5 cm².

Δ LMN is equilateral.

Consider points X, Y and Z as shown.

(1) Side of equilateral triangle = LM = 14 cm.

Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) x side²

∴ A(Δ LMN)= \(\frac{\sqrt{3}}{4}\) x 14² = \(\frac{\sqrt{3}}{4}\)  x 196

∴ A(Δ LMN) = 1.732 × 49     …(as \(\sqrt{3}\) = 1.732)

∴ A(Δ LMN) = 84.868

∴ A(Δ LMN) ~ 84.87 cm²

(2) For a sector L−XZ,

radius (r) = 7 cm

Angular measure of arc (θ) = m(arc XZ) = ∠ XLZ = 60°    ……(Angle of equilateral triangle)

Area of the sector L−XZ = \(\frac{θ}{360}\) × πr² = \(\frac{60}{360}\) × \(\frac{22}{7}\) × 7² =  25.67 cm²

(3) Total area of 3 sectors =3 x Area of the sector L−XZ = 3 × 25.67 = 77.01 cm²

(4) Area of the shaded portion = A(Δ LMN) − Total area of 3 sectors

= 84.87 − 77.01 = 7.86 cm²

Answer is :  (1) A(Δ LMN) = 84.87 cm²,

(2) Area of one of the sectors is 25.67 cm²,

(3) Total area of 3 sectors is 77.01 cm²,

(4) Area of the shaded portion is 7.86 cm².

As per data given in question, fig is as below.

The radius (r) of the circle = AC = 7  cm.

In Δ ABC,

seg AB ≅ seg AC      …(Radii of the same circle)

∴ ∠ ABC ≅ ∠ ACB     ... (Isosceles triangle theorem)

∠ ABC = ∠ ACB = 45°

In Δ ABC,

∠ ABC + ∠ ACB + ∠ BAC =180°   ... (Sum of all angles of a triangle is 180°)

∴ 45° + 45° + ∠BAC = 180°

∴ 90° + ∠ BAC = 180°

∴ ∠ BAC =180° − 90°

∴ ∠ BAC = 90°

m(arc BXC) = ∠ BAC = θ = 90°   ... (Definition of measure of minor arc)

A(segment BXC) = r² \([\frac{πθ}{360}-\frac{sin\,θ}{2}]\)

= \((7\sqrt{2})^2[\frac{3.14×90}{360}-\frac{sin\,90}{2}]\)

= 98 × \([\frac{3.14}{4}-\frac{1}{2}]\)      …. ( sin 90° = 1)

= 98 × \(\frac{1.14}{4}\) = 27.93 cm²

Answer is :  A(segment BXC) = 27.93 cm²

[Note : In the question, if ∠ ABC =45°, then the area of segment BXC is 27.93 cm² and not 3.72 cm² as given in the textbook but if we consider ∠ BAC = 45° instead of ∠ ABC = 45° then area of segment BXC will be 3.92 cm² which is near to 3.72 cm² as given in the textbook, if ∠ BAC = 45° then solution would be as follows.]

The radius (r) of the circle = AC = 7  cm.

m(arc BXC) = ∠ BAC = θ = 45°

A(segment BXC) = r²\([\frac{πθ}{360}-\frac{sin\,θ}{2}]\)

= \((7\sqrt{2})^2[\frac{3.14×45}{360}-\frac{sin\,45}{2}]\)

= 98 × \([\frac{3.14}{8}-\frac{1}{2\sqrt{2}}]\)     …. ( sin 45° = \(\frac{1}{\sqrt{2}}\) )

= 98 × \([\frac{0.32}{8}]\) = 3.92 cm²

Answer is :  A(segment BXC) = 3.92 cm²

The radius (r) of the circle = OP =10 cm.

m(arc PQR) = θ = 60°.

The shaded region is segment PQR,

A (segment PQR) = r²\([\frac{πθ}{360}-\frac{sin\,θ}{2}]\)

= 10²\([\frac{3.14×60}{360}-\frac{sin\,60}{2}]\)

= 10²\([\frac{3.14}{6}-\frac{\sqrt{3}}{2×2}]\)    …. ( sin 60° = \(\frac{\sqrt{3}}{2}\))

= 100 × \(\frac{1.09}{12}\)           … …(as \(\sqrt{3}\) = 1.732)

= \(\frac{109}{12}\) = 9.08 cm²

Answer is : Area of the shaded region is 9.08 cm².

The radius (r) of the circle = AP =7.5.

m(arc PQR) = ∠ PAR = θ =30°.

Area of the segment PQR = r²\([\frac{πθ}{360}-\frac{sin\,θ}{2}]\) = 7.5²\([\frac{3.14×30}{360}-\frac{sin\,30}{2}]\)

= 7.5²\([\frac{3.14}{12}-\frac{1}{2×2}]\)       …. ( sin 30° = \(\frac{1}{2}\))

= 7.5² \([\frac{0.14}{12}]\) = \(\frac{7.785}{12}\) = 0.65625 sq units

Answer is : Area of segment PQR is 0.65625 sq units.

Area of the shaded region is 114 cm2.

The shaded region is segment PRQ.

A (segment PRQ) = 114 cm²

m (arc PRQ) = ∠ POQ = θ = 90°

A (segment PRQ) = r²\([\frac{πθ}{360}-\frac{sin\,θ}{2}]\) 

∴ 114 = r² \([\frac{3.14×90}{360}-\frac{sin\,90}{2}]\)

∴ 114 = r² \([\frac{3.14}{4}-\frac{1}{2}]\)

∴ 114 = r² \([\frac{1.14}{4}]\)

∴ r² = \(\frac{114×4}{1.14}\) = 400

∴ r = 20 cm

Answer is : The radius of the circle is 20 cm.

Let the centre of the circle be O.

Consider points M and N as shown. The radius of

the circle (r) = 15 cm.

m(arc PMQ) = ∠ POQ = θ = 60°

Area of the minor segment = r²\([\frac{πθ}{360}-\frac{sin\,θ}{2}]\) = 15² \([\frac{3.14×60}{360}-\frac{sin\,60}{2}]\)

= 225\([\frac{3.14}{6}-\frac{\sqrt{3}}{2×2}]\) = 225\([\frac{3.14}{6}-\frac{1.73}{4}]\) = 225\([\frac{1.09}{12}]\) = 20.43 cm²

A(circle) = πr²

= 3.14 × 15 × 15

= 3.14 × 225

= 706.5 cm²

A(major segment) = A(circle) − A(minor segment)

= 706.5 − 20.43

= 686.07 cm²

Answer is : A(minor segment) is 20.43 cm² and A (major segment) is 686.07 cm².