Mensuration

Radius of base of cone, r = 1.5cm

Perpendicular height of cone, H = 5cm

As we know that,

Volume of the cone,

On substituting the given values,

⇒ V = 11.79 cm³

∴ Volume of the cone is 11.79 cm³

Diameter of the sphere, d = 6 cm

⇒ Radius of sphere, r = 3 cm

As we know, the volume of sphere,

⇒ V = 113.04 cm³

∴ Volume of sphere is 113.04 cm³

Radius of base of cylinder, r = 5 cm

Height of cylinder, H = 40 cm

As we know,

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

On substituting the values, we get,

A = 2× (3.14)× 5× (5 + 40)

⇒ A = 1413 sq.cm

∴ Total surface area of square is 1413 sq. cm

Radius of sphere, r = 7cm

As we know, the surface area of sphere, A = 4πr²

On substituting the values, we get,

⇒ A = 616 sq. cm

∴ Surface area of sphere is 616 sq. cm

Since the volume of cuboid = length× breadth× height

⇒ Volume of cuboid = Product of the given three dimensions

⇒ Volume of cuboid = 44× 21× 12

Height of cone, H = 24 cm

Let Radius of cone be r

Volume of cone

As the cone is melted to form a cone,

⇒ Volume of cone = volume of cuboid

⇒ r = 21 cm

∴ The radius of the cone is 21 cm

Height of water jug, H_J = 10cm

Radius of water jug, R_J = 3.5 cm

Volume of conical jug,

Let the number of jugs be n.

Height of cylindrical pot, H_P = 10 cm

Radius of pot, R_P = 7 cm

Since the water is transferred from pot to ‘n’ number of jugs,

⇒ Volume of pot = n × Volume of jug

On substituting the given values,

⇒ n = 3× 2²× 1

⇒ n = 12 cm

∴ The cylindrical pot can hold 12 jugs of water.

Let the radius of base be r.

Let the height of cone = H

Height of cylinder, h = 3cm

Area of base, A = 100 sq. cm

As we know the area of circle is πr²

⇒ πr² = 100 ….. (1)

Volume of complete solid figure,V = Volume of cone + volume of cylinder

It is given that volume of solid figure, V = 500 cubic cm

On substituting the value of V and πr² from eq (1), we get,

⇒ H = 6 cm

Total height of figure = h + H = 3 + 6 = 9 cm

∴ total height is 9 cm

Radius of circular region, r = 3cm

Height of Cylinder, H = 40cm

Height of cone, h = 4cm

As we know,

Surface area of hemisphere, A_H = 2πr²

Surface area of cylinder, A_cy = 2πrH

Surface area of cone is,

A_CO = 5πr

Total area = A_H + A_CY + A_CO

On substituting the values, we get,

⇒ Total area = πr(2r + 2H + 5)

⇒ Total area = π× 3× (6 + 80 + 5)

⇒ Total area = 273π

∴ total area of the figure is 273π sq. cm

Since there are only flat tablets which are stacked onto each other inside the cylindrical wrapper, so the radius of the wrapper does not matter.

Height of wrapper, H = 10cm = 100mm

Thickness of the tablet, h = 5mm

⇒ Number of the tablets = 20

∴ There are 20 tablets inside the cylindrical wrapper.

Radius of the circular base, r = 3cm

Height of cone, h = 4cm

⇒ V =3.14×3 × 10

⇒ V = 94.20 cubic cm

Now,

⇒ S_C = 3.14 × 3 × 5

⇒ S_C = 47.1 sq.cm

Surface area of sphere, S_S = 2πr²

⇒ S_S = 2× (3.14) × 3²

⇒ S_S = 56.52 sq.cm

Surface area of toy = 47.1 + 56.52 = 103.52 sq.cm

∴ Volume and surface area of toy is 94.20 cm³ and 103.52 cm² respectively

Diameter of spherical ball, d = 42 cm

Radius of ball, r =21 cm

As we know that, Surface area of sphere, A = 4πr²

⇒ A = 4× (3.14) × (21)²

⇒ A = 5538.96 sq.cm

Also,

⇒ V = 38772.72 cm³

∴ The surface area and volume of the spherical ball is 5538.96 cm² and 38772.72 cm³ respectively

Diameter of glass, D = 14 cm

⇒ Radius of glass, r = 7cm

Height of water level, H = 30 cm

We know that, volume of cylinder, V = πR²H

⇒ V = π × (7)²× 30

⇒ V = 1470π cubic cm

Diameter pof metal sphere, d = 2 cm

Radius of metal sphere, r = 1cm

As we know that,

⇒ v = 1.33π cubic cm

Volume of water = Volume of glass – Volume of metal sphere

⇒ Volume of water = 1470π – 1.33π

∴ volume of the water is 1468.67π cm³

The two radii of frustum are, r₁ = 14cm and r₂ = 7cm

Height of bucket, H = 30 cm

As we know that,

⇒ V = 10780 cm³

Now, as 1litre = 1000cm³

⇒ V = 10.780 litre

∴ Bucket can hold 10.780 litres of water

(i) The two radii of frustum are, r₁ = 14cm and r₂= 6cm

Height of frustum, H = 6cm

Slant height of frustum,

⇒ l = 10

As we know that,

Curved surface area, A_C = πl(r₁ + r₂)

⇒ A_C = (3.14) × 10 × (14 + 6)

⇒ A_C = (3.14) × 200

⇒ A_C = 628 sq. cm

∴ the curved surface area of frustum is 628 sq. cm

(ii) Total surface area, A_T = Curved surface area + area of the two circular regions

A_T = A_C + πr₁² + πr₂²

On substituting the above values, we get,

⇒ A_T = 628 + (3.14) × (14² + 6²)

⇒ A_T = 628 + (3.14)× (196 + 36)

⇒ A_T = 628 + 728.48

⇒ A_T = 1356.48 sq. cm

∴ the total surface area of frustum is 1356.48 cm²

(iii) As we know,

On substituting the values, we get,

V = 1984.48 cm³

∴ Volume of the frustum is 1984.48 cm³

Circumference₁ = 2πr₁ = 132

Circumference₂ = 2πr₂ = 88

Slant height of frustum,

Curved Surface area of frustum = π(r₁ + r₂)l

⇒ Curved Surface Area of frustum = π× (21 + 14) × 25

⇒ Curved surface area = π × 35 × 25 = 2750 sq. cm

Radius of circle, r = 10 cm

Angle made between the arc ,θ = 54°

⇒ Area of sector = 47.1 sq. cm

Measure of an arc of circle, θ = 80°

Radius of circle, r = 18 cm

⇒ Area of sector = 25.12 cm

Radius of circle, r = 3.5 cm

Length of arc, l = 2.2 cm

As we know,

Also,

⇒ Area of sector = 3.85 sq. cm

Radius of circle, r = 10 cm

Area of sector (minor sector) = 100 sq. cm

Area of circle, A_C = πr²

⇒ A_C = 3.14× 10²

⇒ A_C = 314 sq. cm

Area of major sector, A_M = area of circle – area of minor sector

⇒ A_M= A_C – 100

⇒ A_M = 314 -100 = 214 sq. cm

∴ area of major sector is 214 sq. cm

Radius of circle, r = 15 cm

Also,

On substituting the values, we get,

⇒ Length of arc = 4 cm

(1) Radius of circle, r = 7cm

Area of circle,A_C = πr²

⇒ A_C = 154 sq. cm

∴ area of circle is 154 sq.cm

(2) Angle subtended by the arc = 60°

As we know,

⇒ A(O- MBN) = 25.7 sq. cm

(3) A(O- MCN) = Area of circle – A(O – MBN)

⇒ Area(O – MCN) = A_C – 25.7

⇒ Area(O – MCN) = 154 – 25.7

∴ Area of major sector is 128.3 sq. cm

Radius of circle, r = 3.4 cm

Perimeter of sector, P = 12.8

⇒ P = length of arc + 2× radius

⇒ Length of arc, l = P – 2×r

⇒ l = 12.8 – 2(3.4)

⇒ l = 6 cm

Let the ∠ APC be θ

As we know that,

Also ,

On Substituting the value of theta from above equation,

⇒ A = 10.2 sq. cm

∴ Area of Sector is 10.2 sq. cm

Let the ∠ ROQ= ∠ MON = θ = 60°

As we know that,

⇒ Length(RXQ) = 7.6 cm

Similarly,

⇒ Length (MYN) = 22 cm

As we know that,

(1) Let the ∠APC be θ

Radius of circle, r =14 cm

Area of sector, A = 154 cm²

⇒ θ = 90°

(2) Since the angle formed is 90° , which is one–fourth of the perimeter of circle

⇒ l = 22 cm

Radius of circle, r = 7cm

(l) Angle subtended by arc, θ = 30°

As we know,

⇒ A_l = 12. 83 sq. cm

(ll) Angle subtended by arc, θ = 210°

Similarly,

⇒ A_ll = 89.83 sq. cm

(lll) Angle subtended by arc, θ = (3× 90)° = 270°

Similarly,

⇒ A_lll = 115.5 sq. cm

As we know that,

Given area of sector, A = 3.85 sq.cm

Radius of circle= r

Central angle, θ = 36°

On substituting the values, we get,

⇒ r = 3.5 cm

∴ Radius of circle is 3.5 cm

Since part x is a sector of a circle with radius, r = 14 cm and the central angle is 90°, so the area of x will be equal to one-fourth of the area of circle with PQ as radius.

Area of circle with PQ as radius = π (PQ)²

⇒ x = 154 sq. cm

Similarly, area y is also equal to one-fourth od area of circle with radius, r = QR – PQ

⇒ r = 21 – 14 = 7 cm

Area of circle with r as radius = π (r)²

⇒ y = 38.5 sq. cm

Also,

z = Area of rectangle(PQRS) – x – y

Area of rectangle = PQ× QR

⇒ Area of rectangle = 14× 21 = 294 sq. cm

⇒ z = 294 – 154 – 38.5

⇒ z = 101.5 sq.cm

∴ the area of x, y and z are 154 sq.cm, 38.5 sq.cm and 101.5 sq.cm respectively

(1) Side of triangle = LM = a = 14 cm

Since Δ LMN is an equilateral triangle, so the area of the triangle is given by:

⇒ A_T = 84.87 sq.cm

(2) Angle subtended by the corner = θ = 60°

As we know,

Here ,

⇒ A_S = 25.67 sq. cm

(3) Total area of all sector, A_TS = 3× A_S

⇒ A_TS = 3× 25.67

⇒ A_TS = 77.01 sq.cm

(4) Area of shaded region, A_R = Area of triangle – Area of all three sectors

⇒ A_S = A_T - A_TS

⇒ A_S = 84.87 – 77.01

⇒ A_S = 7.86 sq. cm

∴ area of shaded region is 7.86 sq. cm

From the property, we know that, If two sides of a triangle are equal then their corresponding angles are also equal.

So, as AB = AC,

⇒ ∠ ABC = ∠ ACB = 45°

As the sum of angles of a triangle is equal to 180°

⇒ ∠ BAC = 90°

⇒ A_T = 49 sq.cm

Area of the sector = one-fourth of a circle

⇒ A_S = 77 sq.cm

Area of the shaded region, A_R = A_S - A_T

⇒ A_R = 77 – 49

⇒ A_R = 28 sq.cm

∴ The area of the shaded region is 28 sq. cm.

Since the angle subtended at centre is 60°

And by the property, if two sides of a triangle are equal then their corresponding angles are also equal.

⇒ ∠ ORP = ∠ OPR

As the sum of all internal angles of a triangle is equal to 180°

⇒ ∠ ORP = ∠ OPR = 60°

⇒ Δ OPR is an equilateral triangle.

⇒ A_T = 43.25 sq. cm

Area Of Sector (O-PQR), A_S is given as:

⇒ A_{S =} 52.33 sq.cm

Area of shaded region, A_R = A_s – A_T

⇒ A_R = 52.33 – 43.25

⇒ A_R = 9.08 sq.cm

∴ Area of shaded region is 9.08 sq.cm

Radius of circle, r = 7.5 cm

∠ PAR = θ = 30°

Area(A – PQR), A_S:

⇒ A_S = 14.71 sq.cm

Also,

⇒ A_T = 14.06 sq. cm

Area of segment PQR, A_R = A_S - A_T

⇒ A_R = 14.71 – 14.06 = 0.6562 sq.cm

∴ Area of shaded region is 0.6562 sq.cm

Area Of shaded region, A_R = 114 sq.cm

Area of sector (O-PRQ),A_S = one-fourth of area of circle

Area of Shaded Region, A_R = A_S – A_T

⇒ r = 20 cm

∴ Radius of the circle is 20 cm

Radius of circle, r = 15cm

Central angle, θ = 60°

Since the angle subtended at centre is 60°

And by the property, if two sides of a triangle are equal then their corresponding angles are also equal.

⇒ ∠ OQP = ∠ OPQ

As the sum of all internal angles of a triangle is equal to 180°

⇒ ∠ OQP = ∠ OPQ = 60°

⇒ Δ OPQ is an equilateral triangle.

A_T = 97.32 sq.cm

⇒ A_R = 117.75 – 97.32

⇒ A_R = 20.43 sq.cm

Now,

⇒ A_S = 706.5 – 20.43

⇒ A_S = 686.07 sq.cm

∴ The area of minor segment and major segment is 20.43 sq.cm and 686.07 sq.cm respectively

Circumference of circle, C = 2πr

Area Of circle, A = πr²

⇒ r = 7

Circumference = 2× r× π

Circumference = 14π

∴ Option (A) is correct

Measure of arc, θ = 160°

Length of arc, l = 44cm

As we know that,

⇒ Circumference = 99 cm

∴ Circumference of circle is 99cm, Option (D) is correct answer.

Central angle, θ = 90°

Radius, r = 7cm

As we know that,

⇒ l = 11cm

Perimeter of sector =Length of sector + (2× Radius)

Perimeter = 11 + (2 × 7)

Perimeter = 25cm

∴ Option (B) is correct.

Radius, r = 7cm

Vertical Height, h = 24 cm

⇒ l = 25 cm

Curved Surface Area of cone, A = πrl

⇒ A = 550 sq.cm

∴ Option (B) is correct.

Radius of cylinder, r = 5cm

Height of cylinder = h

Curved Surface Area of cylinder, A = 440 cm² = 2πrh

440 = 2× π × 5 × h

∴ Option (A) is correct.

Height of cylinder, H = 5 cm

Height of cone =h

Volume of cylinder, V = πr²H

Since cone is melted to form cylinder,

⇒ V = v

⇒ h = 3× H

⇒ h = 3 × 5

⇒ h = 15 cm

∴ Option (A) is correct

Side of the cube, a = 0.01 cm

Volume of cube, V = a³

⇒ V = (0.01)³

⇒ V = 0.000001 cm³

∴ Option (D) is correct

Let the side of cube be ‘a’ cm

Volume of cube, V = a³

⇒ As 1m³ = 10⁶ cm3

⇒ a³ = 10⁶

⇒ a = 100 cm

∴ Option (B) is correct

Height of frustum, h = 21 cm

Two radii of frustum are, r₁ = 20 cm and r₂ = 15 cm

As we know,

⇒ V = 20350 cm³

⇒ V = 20.35 litres

∴ Capacity of bucket is 20.35 litres

Volume of 1 plastic ball = V_B

Radius of ball, r_B = 1 cm

Length of tube, h = 90 cm

Outer radius of tube, r_O = 30 cm

Thickness of tube = 2 cm

Inner radius, r_I = 30 – 2 = 28 cm

Volume of tube, V_T = Volume of Outer cylinder – Volume of inner cylinder

⇒ V_T = 90π(30² - 28²)

⇒ V_T = 90× 116 × π

Let ‘n’ be the number of balls melted.

⇒ V_T = n × V_B

⇒ n = 7830 balls

∴ 7830 balls were melted to make the tube.

Volume of parallelepiped, V_P = Product of its three dimensions

⇒ V_P = 16 × 11 × 10

⇒ V_P = 1760 cm³

Thickness of coin, t = 2mm = 0.2 cm

Diameter of coin = 2 cm

⇒ Radius of Coin, r = 1 cm

Volume of one coin, V_C = πr²t

⇒ V_C = 0.6285 cubic cm

Let the number of coins made be ‘n’

⇒ V_P = n × V_C

⇒ n = 2800

∴ 2800 coins are made.

Diameter of roller = 120 cm

⇒ Radius, r = 60 cm

Length Of Roller, h = 84 cm

⇒ A_R = 31680 sq. cm

Total area of ground, A_G = 200 × A_R

⇒ A_G = 6336000 cm²

As 1 m² = 10⁴ cm²

⇒ A_G = 633.6 sq.m

Rate to level the ground = Rs. 10 per sq.m

⇒ Total Expenditure = 633.6 × 10

⇒ Total expenditure = Rs. 6336

∴ Total expenditure to level the road will be Rs. 6336

Thickness of sphere, t = 0.01m = 1 cm

Diameter of sphere = 12 cm

⇒ Outer Radius of sphere, R_O = 6 cm

⇒ Inner Radius of sphere, R_I = R_O – t = 6 – 1 = 5 cm

Outer Surface area of sphere, A = 4πR_O²

⇒ A = 4× 3.14 × 6²

⇒ A = 452.16 sq.cm

⇒ V = 381.33 cm³

⇒ Mass = 8.88 × 381.33

⇒ Mass = 3385.94 gm

∴ Outer surface area and mass of the sphere is 452.16 sq.cm and 3385.94 gm

Diameter of cylinder = 28 cm

⇒ Radius of cylinder, R = 14 cm

Height of cylinder, H = 20 cm

Volume of cylinder, V_C = πR²H

Height of cone, h = 14 cm

Radius of cone = r

Now, as the sand in cylinder forms the cone,

⇒ V = v

⇒ r = 28.98 cm

Base area = πr²

⇒ Base area = 2640 sq. cm

∴ The base area of the cone formed is 2640 sq. cm

Radius of sphere, R = 9 cm

Let the length of wire be H

Radius Of wire, r = 2 mm = 0.2 cm

Volume of wire , v = πr²H

As sphere is melted to make a wire,

⇒ V = v

⇒ H = 24300 cm

⇒ H = 243 m

∴ The length of the wire formed is 243 m

Let the measure of arc be θ

Area of sector = 15π

Radius, r = 6 cm

As we know,

⇒ θ = 150°

Also,

⇒ l = 5π

∴ The length of the arc is equal to 5π.

Let ∠ APB = θ

Therefore,

⇒ θ = 120°

⇒ A_T = 27.68 sq.cm

Also,

⇒ A_S = 66.96 sq.cm

Area of shaded region, A_R = A_S - A_T

⇒ A_R = 66.96 – 27.68

⇒ A_R = 39.28 sq. cm

∴ area of shaded region is 39.28 sq. cm

Side of square ABCD = radius of sector C - BXD = 20 cm

Area of square = (side)² = 200² = 400 sq.cm

Area of shaded region inside the square = Area of square ABCD - Area of sector C - BXD

⇒ 76 sq.cm

Radius of bigger sector = Length of diagonal of square ABCD

Area of the shaded regions outside the square = Area of sector A - PCQ - Area of square ABCD = A(A - PCQ) - A(ABCD)

⇒ 628 – 400

⇒ 228 sq.cm

∴ total area of the shaded region = 86 + 228 = 314 sq.cm.

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

AQ = 2r = AB – BQ

2r = 50 - 9 = 41