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Correct option is: A) 5

Let the radius of third sphere be r cm.

\(\because\) Three spheres of radii 3 cm, 4 cm and r cm are made by melting a solid metallic sphere of radius 6 cm.

\(\therefore\) Volume of sphere of 6 cm radius = Combined volume of formed three spheres

\(\Rightarrow\) \(\frac 43 \pi \times 6^3 = \frac 43 \pi \times 3^3 + \frac 43 \pi \times 4^3 + \frac 43 \pi r^3 \)

\(\Rightarrow\) \(6^3 = 3^3 + 4^3 +r^3\) 

= \(r^3 = 6^3 - 3^3 - 4^3 = 216 - 27 - 64\) 

= 216 - 91 = 125 = \(5^3\) 

= r = 5 cm 

Hence, the radius of the third sphere is 5 cm.

Correct option is: A) 5

Correct option is: C) 5

Given that height of the cone is h = 4 cm and radius of the cone is r = 3 cm.

\(\because\) \(l^2 = r^2 + h^2\)

= \(3^2+4^2 = 9+ 16 = 25 = 5^2\)

\(\because\) l = 5 cm

Hence, the slant height of the cone is 5 cm.

Correct option is: C) 5

Correct option is: B) 314

Given that the area of the base of right circular cone = 78.5 \(cm^2\)

\(\therefore\) \(\pi r^2\) =  78.5 \(cm^2\)

Also given that height of the cone is h = 12 cm.

\(\therefore\) Volume of cone = \(\frac 13 \pi r^2h\)

= \(\frac 13 \times 78.5 \times 12\)

= 78.5 \(\times\) 4

 = 314 \(cm^3\)

Hence, the volume of the cone is 314 \(cm^3\).

Correct option is: B) 314

l = 12 cm 

T.S.A of cube = 6l² = 6 x 12²= 6 x 144 = 864 cm² 

Volume of cube = P = (12)³ = 1728 cm³

The cost of Rs. 500 is for 1m³ . 

The cost of Rs. 256 is for 

\(\frac{256}{500} = \frac{128}{250}\) = \(\frac{64}{125}m^3\)

∴ Volume of the wood = \(\frac{64}{125}m^3\) = 0.512 m³ 

Volume = 0.512m 

l³ = 0.512 

l = \(3\sqrt{0.512}\)

l = 0.8 m² 

= 0.8 × 100 

Length of the side = 80 cm, l = 80 cm

T.S.A of a cube = 486 cm² 

6¹² = 486 

l² = \(\frac{486}{6}\)

l² = 81 

∴ l = √81 = 9 cm 

Volume=l³ = 9³= 729 cm³

= 10m 

b = 8 

h = 28m 

Volume of soil = volume of cuboid 

= l × b × h = 10 × 8 × 28 = 2240 m³ 

2240 m³ of soil has to be excavated. 

Area to be plastered = L.S.A of cuboid 

= 2h(l + b) 

= 2 ×28(10 + 8) 

= 56 × 18 = 1008m² 

The cost of plastering 1m² = Rs. 15 

∴The cost of plastering = 15 × 1008 = Rs. 15,120

V = 6.4m³ 

l = 2m 

b = 1.6m 

h = ? 

Volume of cuboid = 6.4m³ 

l × b × h = 6.4 

2 × l.6 × h = 6.4 

3.2h = 6.4 

h = \(\frac{6.4}{3.2}\)

h = 2m 

h = 2m . 

∴ The depth of the tank is 2m.

Given,

Ratio of radius of two cylinder = 2 : 3

= \(\frac{r_1}{r_2}\) = \(\frac{2}{3}\)

Ratio of their heights = 5 : 3

= \(\frac{volume\,of\,cylinder\,1}{volume\,of\,cylinder\,2}\) = \(\frac{πr^2_1h_1}{πr^2_2h_2}\) = \(\frac{4\times 5}{9\times 3}\) = \(\frac{20}{27}\)

∴ Ratio of volumes of two cylinder = 20 : 27

Given,

Total surface area of cylinder = 616 cm²

Ratio between curved surface area and total surface area of cylinder = 1 : 2

= \(\frac{2πrh}{2πr(h+r)}\) = \(\frac{1}{2}\)

= \(\frac{h}{h+r}\) = \(\frac{1}{2}\)

= 2h = h + r

= h = r

= 2πr(h + r) = 616 Given

= 2πr x 2r = 616(putting h = r)

= r² = \(\frac{616}{4π}\) = \(\frac{616\times 7}{4\times 22}\) = 49

= r = \(\sqrt{49}\) = 7

Radius = 7 cm = height

∴ Volume of cylinder = πr²h = \(\frac{22}{7}\)x 7 x 7 x 7 = 1078 cm³

Given,

Ratio between radius and height of a cylinder = 2:3

= \(\frac{r}{h}\) = \(\frac{2}{3}\)

= 3r = 2h or h \(\frac{3}{2}\)r

Volume of cylinder = 1617 cm³

= πr²h = 1617

= \(\frac{22}{7}\)x r² x \(\frac{3}{2}\)r = 1617

= r³ = \(\frac{1617\times 14}{66}\) = 343

= r = \(\sqrt{343}\) = 7 cm

Radius = 7 cm

Height = \(\frac{3}{2}\) x 7 = \(\frac{21}{2}\) cm = 10.5 cm

∴ total surface area of cylinder = 2πr(h + +r) = 2 x \(\frac{22}{7}\) x 7(7 + 10.5) = 44 x 17.5 = 770 cm²

Given,

= \(\frac{curved\,surface\,area\,of\,cylinder}{total\,surface\,area\,of\,cylinder}\) = \(\frac{1}{2}\)

Let radius of cylinder = r

Let height of cylinder = h

So,

= \(\frac{2πrh}{2πr(h+r)}\) = \(\frac{1}{2}\)

= \(\frac{h}{h+r}\) = \(\frac{1}{2}\)

= 2h = h + r

= 2h - h = r

= h = r, height = radius

Given,

Curved surface area = 1320 cm²

Diameter of base = 21 cm

Radius of base = \(\frac{diameter}{2}\) = \(\frac{21}{2}\) cm

So,

= 2πrh = 1320

= 2 x \(\frac{22}{7}\) x \(\frac{21}{2}\) x h = 1320

= h = \(\frac{1320\times 14}{2\times 22\times 21}\) = 20 cm

∴ Volume of cylinder = πr²h = \(\frac{22}{7}\)x \(\frac{21}{2}\) x \(\frac{21}{2}\)x 20 = 6930 cm³

Given,

Curved surface area of cylinder = 1320 cm²

Diameter of base = 21 cm

Radius of cylinder = \(\frac{21}{2}\) cm

Let height of cylinder = h cm

So,

= 2πrh = 1320

= h = \(\frac{1320}{2πr}\) = \(\frac{1320\times 7\times 2}{2\times 22\times 21}\) = 20 cm

Height of cylinder = 20 cm

Area of rhombus = 1/2 (Product of its diagonals)

Area of each tile = (1/2 x 45 x30) cm²

= 675 cm² 

Area of 3000 tiles = (675 × 3000) cm² 

= 2025000 cm² = 202.5 m² 

The cost of polishing is Rs 4 per m^2.

Cost of polishing 202.5 m² area 

= Rs (4 × 202.5)  = Rs 810 

Thus, the cost of polishing the floor is Rs 810.

It is given that diameter of the (oil drum) cylinder = 2 m. 

Radius of cylinder = \(\frac{d}{2}=\frac{2}{2}\) = 1 m. 

Total surface area of a cylindrical drum 

= 2πr(r + h) 

= 2 × \(\frac{22}{7}\) × 1(1 + 7) 

= 2 × \(\frac{22}{7}\) × 8 = \(\frac{352}{7}\)m² = 50.28 m² . 

So, the total surface area of a drum = 50.28 m² 

Painting charge per 1 m² = ₹ 5. 

Cost of painting of 10 drums

= 50.28 × 5 × 10 

= ₹ 2514

(i) We have,

Edge of cube = 1.2 m

Surface area of cube = 6 x side² = 6 x 1.2² = 6 x 1.44 = 8.64 m²

(ii) We have,

Edge of cube = 27 cm

Surface area of cube = 6 x side² = 6 x 27² = 6 x 729 = 4374 cm²

(iii) We have,

Edge of cube = 3 cm

Surface area of cube = 6 x side² = 6 x 9 = 54 cm²

(iv) We have,

Edge of cube = 6 m

Surface area of cube = 6 x side² = 6 x 6² = 216 m²

(v) We have,

Edge of cube = 2.1 m

Surface area of cube = 6 x side² = 6 x 4.41 = 26.46 m²

(i) Given,

Length = 10 cm

Breadth = 12 cm

Height = 14 cm

So,

Surface area of cuboid = 2(lb x bh x hl) = 2(10 x 12 + 12 x 14 + 14 x 10)

= 2(120 + 168 + 140) = 2 x 428 = 856 cm²

(ii) Given,

Length = 6 dm

Breadth = 8 dm

Height = 10 dm

So,

Surface area of cuboid = 2(lb x bh x lh) = 2(6 x 8 + 8 x 10 + 10 x 6)

2(48 + 80 + 60) = 2 x 188 = 376 dm² 

(iii) Given,

Length = 2 m

Breadth = 4 m

Height = 5 m

So,

Surface area of cuboid = 2(lb x bh x lh) = 2(2 x 4 + 4 x 5 + 5 x 2)

= 2(8 + 20 + 10) = 2 x 38 = 76 m²

(iv) Given,

length = 3.2 m= 32 dm

breadth = 30 dm

height = 250 cm= 25 dm

so,

surface area of cuboid = 2(lb x bh x lh) = 2(32 x 30 + 30 x 25 + 25 x 32)

= 2(960 + 750 + 800) = 2 x 2510 = 5020 dm²

Given,

Dimensions of solid rectangular piece = 6 m × 6 cm ×2 cm

Volume of rectangular iron = 600 cm x 6 cm x 2 cm = 7200 cm³

Weight of 1 cm³ iron = 8 gm

∴ weight of 7200 cm³ = 7200 x 8 = 57600 gm = 57.6 kg

(i) 1 m³ = 1 x (100 x 100 x 100) = 10⁶ cm³ [1 m = 100 cm]

(ii) 1 litre = 1000 cm³ = 1000 x (0.1 x 0.1 x 0.1) dm³ = 1 dm³ [1 cm = 0.1 dm]

(iii) 1 kl = 1000 litre = 1 m³ [1 m³ = 1000 litre]

(iv) Side of cube = 8 cm

Volume of cube = 8³ = 512 cm³

(v) Volume of cuboid = 4000 cm³

Length = 10 cm, breadth = 8 cm

Then,

Height = \(\frac{volume}{length\times breadth}\) = \(\frac{4000}{10\times 8}\) = 50 cm

(vi) 1 cu. dm = 1 dm³ = 1 x 10 x 10 x 10 = 10³ cm³[1 dm = 10 cm]

= 10³ x 10 x 10 x 10 x 10 mm³ = 10⁶ mm³ [1 cm = 10 mm]

(vii) 1 km³ = 1000 x 1000 x 1000 = 10⁹ m³ [1 km = 1000 m]

(viii) 1 litre = 1000 cm³ = 10³ = cm³

(ix) 1 ml = \(\frac{1}{1000}\)litre = \(\frac{1}{1000}\) x 1000 = 1 cm³ [1 ml = \(\frac{1}{1000}\) litre]

(x) 1 kl = 1 x 1000 litre =1 m³ = 1 x (10 x 10 x 10) dm³ [1m = 10 dm]

1 kl = 10³ dm³ = 1000 x 1000 = 10⁶ cm³

Area of rectangle = length × Breadth

(a) l = 9 m b = 6 m

Area = l × b = 9 × 6 = 54 m²

(b) l = 17 m b = 3 m

Area = l × b = 17 × 3 = 51 m²

(c) l = 4 m b = 14 m

Area = l × b = 4 × 14 = 56 m²

It can be see that rectangle (c) has the largest area and rectangle (b) has the smallest area.

It’s known that, Area o f square = side × side

(a) 10cm

side = 10 cm

Area of the square =10 × 10= 100 cm²

(b) 14 cm

side = 14 cm

Area of the square = 14 × 14 = 196 cm²

(c) 5m

Side = 5 m

Area of the square = 5 × 5 = 25 m²

It is known that

Area of rectangle = length × Breadth

(a) 1 = 3 cm, b = 4 cm 

Area = l × b = 3 × 4 = 12 cm²

(b) 1 = 12 m, b = 21 m

Area = l × b = 12 × 21 = 252 m²

(c) l = 2 km, b = 3 km

Area = l × b = 2 × 3 = 6km²

(d) l = 2 m , b = 70 cm = 0.70 m

Area = l × b = 2 × 0.70 = 1.40 m²

Given,

Let edge length of three equal cubes = a

Then,

Sum of surface area of 3 cubes = 3 x 6a² = 18a²

When these cubes are placed in a row adjacently they form a cuboid.

Length of new cuboid formed = a + a + a = 3a

Breadth of cuboid = a

Height of cuboid = a

Total surface area of cuboid = 2(lb x bh x hl) = 2(3a x a + a x a + a x 3a)

= 2(3a² + a² + 3a²) = 2 x 7a² = 14a²

Hence,

= \(\frac{total\,surface\,area\,of\,new\,cuboid}{sum\,of\,surface\,area\,of\,3\,cuboids}\) = \(\frac{14}{18}\) = \(\frac{7}{9}\) = 7:9

Given,

Let edge of cube = a

So volume of cube = a³

Case (i)

Edge become = \(\frac{a}{2}\)

Volume become = \((\frac{a}{2})^3\) = \(\frac{a^3}{8}\) = \(\frac{1}{8}\) times

Case (ii)

Edge becomes = 3a

Volume become = \((3a)^3\) = 27a^{3 = 27 times}

Given,

Volume of wood in cuboidal block = 36 cm³

Length of block = 4 cm

Breadth of block = 3 cm

Let height of block = h cm

So,

= l x b x h = 36

= h = \(\frac{36}{4\times 3}\) = 3 cm

Given,

Length of milk container = 8 cm

Width = 50 cm

Volume to hold = 4 litre = 4000 cm³

Let height of container = h cm

So,

= l x b xh = 400

= h = \(\frac{4000}{50\times 8}\) = 10 cm

Area of trapezium = \(\frac{1}{2}\)(sum of lengths of parallel sides)

Length of bases of trapezium = 60 cm and 60 cm

Area of trapezium = \(\frac{1}{2}\)(60 + 60) x altitude

Length of altitude = \(\frac{2\times area}{120}\) = \(\frac{2\times 600}{120}\) = 10

Therefore length of altitude = 10 cm

Area of trapezium = \(\frac{1}{2}\)(sum of lengths of parallel sides) x altitude

Area of trapezium = \(\frac{1}{2}\)(sum of sides) x altitude

Sum of parallel sides = \(\frac{2\times area}{2.8}\) = \(\frac{2\times 4.2}{2.8}\) = 3

Therefore sum of length of parallel sides = 3 m

Area of trapezium = \(\frac{1}{2}\)(sum of lengths of parallel sides) x distance between parallel sides

Area of trapezium = \(\frac{1}{2}\)(sum of sides) x distance between parallel sides

distance between parallel sides = \(\frac{2\times area}{sum\, of\, sides}\) = \(\frac{2\times 960}{80}\) = 24

Therefore distance between parallel sides = 24 cm

Length of parallel sides of trapezium = 1.2 m and 1 m

Distance between parallel sides of trapezium = 0.8 m

Area of trapezium = \(\frac{1}{2}\)(sum of sides) x distance between parallel sides

Area of trapezium = \(\frac{1}{2}\)(1.2 + 1) x 0.8

Area of trapezium = 0.88 m²

Let the length of other parallel side of trapezium = x cm

Length of one parallel side of trapezium = 6 cm

Area of trapezium = 28 cm²

Length of altitude of trapezium = 4 cm

Area of trapezium = \(\frac{1}{2}\)(sum of sides) x distance between parallel sides

28 = \(\frac{1}{2}\)(6 + x) x 4

6 + x = \(\frac{28}{2}\)

x = 14 - 6 = 8

Therefore the length of other parallel side of trapezium = 8 cm

(a) False – Boundry wall is around her house, so she must know the perimeter of the land and not the area.

(b) True – Track is prepared along the boundary of the sports ground.

(a) 15.9 cm. 

(b) 25 sq cm

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

= [ 2 x (22/7) x 7(7 + 3) ] m²

= 440 m²

Thus, 440 m² sheet of metal is required.