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(a)

(b)

We know the, isosceles triangle contains two equal sides.

From the question it is given that, one of the two equal sides is 18cm.

And Perimeter of an isosceles triangle is 50cm

We know that, perimeter of isosceles triangle = sum of all sides

Let us assume the third side be x.

Then,

50 = 18 + 18 + x

50 = 36 + x

x = 50 – 36

x = 14 cm

Therefore, length of third side of isosceles triangle is 14 cm.

The length of outer boundary of the park = (200 + 300 + 80 + 300 + 200 + 260) m
= 1340 m

∴ Cost of fencing the park at the rate of Rs. 20 per metre = Rs. (20 × 1340) = Rs. 26800

Area of the flower bed = (100 × 80) m² = 8000 m²

Now, the cost of manuring the flower bed at the rate of Rs. 50 per square metre

= Rs. (50 × 8000)

= Rs. 400000

Length of display board = 1 m 50 cm = 1.50 m

Breadth of display board = 1 m

Perimeter of one display board

= 2(length + breadth) = 2(1.50 + 1) m

= 2(2.50) m = 5 m

∴ Perimeter of 24 display boards = (24 × 5) m = 120 m

Number of boards will be framed using the aluminium strip of 100 m long = 100/5 = 20

∴ The length of the aluminium strip required to frame the remaining boards

= 120 m – 100 m = 20 m

Radius of Hemisphere, r = 7 cm 

T.S.A of Hemisphere = 3πr² 

= 3 × \(\frac{22}{7}\) × 7 × 7 = 462 (cm)²

Volume of the sphere = \(\frac{4}{3}\)πr³ 

= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 21 × 21 × 21 

= 38,808(cm)³

Given,

Length of hall = 18 m

Width of hall = 12 m

Let height of hall = h metre

Then,

Sum of area of floor & flat roof = l x b+ l x b = 12 x 8 = 432 m²

Sum of area of 4 walls = 2(l x h + b x h) = 2(18h + 12h) = 60 m²

Now,

= 60h = 432 ... ... . .. . ..  given

= h = \(\frac{432}{60}\) =7.2 m

Height of hall = 7.2 metre

Given,

Dimensions of cuboidal block of silver = 9 cm × 4 cm × 3.5 cm

Volume of beads made = 1.5 cm³

So,

Number of beads can be made from cuboidal block = \(\frac{9\times 4\times 3}{1.5}\) = 72 beads

Given,

External dimensions of wooden box = 48 cm × 36 cm × 30 cm

Dimensions of bricks = 6 cm × 3 cm × 0.75 cm

Thickness of wood = 1.5 cm

So,

Internal dimensions of box = 48 - (2 x 1.5)cm + 36 - (2 x 1.5)cm + 30 - (2 x 1.5)cm

= 45 cm ×33 cm× 27 cm

Hence,

Number of bricks can be put in box = \(\frac{internal\,volume\,of\,box}{volume\,of\,one\,brick}\)  = \(\frac{45\times 33\times 27}{6\times 3\times 0.75}\)  = 2970 bricks

Given,

Dimensions of cinema hall are = 100 m × 50 m × 18 m

Where, length = 100 m , breadth = 50 m , height = 18 m

Each person air requirement = 150 m³

Now,

Volume of cinema hall = lbh = 100 x 50 x 18 = 90000 cm³

So,

Number of person can sit in cinema hall = \(\frac{volume\,of\,hall}{volume\,of\,air\,required\,by\,one\,person}\) = \(\frac{90000}{150}\) = 600

Area: 

(i) 11cm² 

(ii) 13cm² 

(iii) 13cm² 

Perimeter: 

(i) 18cm 

(ii) 28cm 

(iii) 28cm

Correct option is: C) 4 : 1

Let the radius & height of the original cylinder be r & h respectively.

\(\therefore\) Volume of original cylinder is \(V_1 = \pi r^2 h\)

When the radius of base of cylinder is halved and height remaining same, then, the volume of formed cylinder is

\(V_2 = \pi (\frac r2)^2h = \frac {\pi r^2h}{4}\)

Now, \(\frac {V_1}{V_2}\) = \(\frac {\pi r^2h}{\frac {\pi r^2h}4}\) = \(\frac 41 = 4:1\)

Hence, the ratio of the volume of cylinder thus obtained to the volume of the original cylinder is 4 :1.

Correct option is: C) 4 : 1

L.S.A of cube = 4l² 

64 = 4l² 

\(\frac{64}{4}\) = l² 

∴ l = √16 = 4m 

The side of the cube is 4m.

l = 4m 

Area to be white washed = L.S.A of cube 

= 4l² = 4 × 4² = 4 × 16 = 64 m² 

cost of white washing 1m² = 20 

∴ The cost of white washing 64m² = 64 × 20 = Rs. 1280.

Correct option is: C) 1848

Given that \(\frac rh = \frac 32 \) and r = 21 cm

\(\therefore\) h = \(\frac 23 r = \frac 23 \times 21 = 14 \, cm\) 

\(\therefore\) LSA of cylinder = \(2 \pi rh\)

= \(2 \times \frac {22}7 \times 21 \times 14 = 44 \times 42= 1848 \, cm^2\)

Hence, the lateral surface area of cylinder is 1848 \(cm^2\)

Correct option is: C) 1848

Correct option is: C) 21

Given that the volume of the cylinder is 49896 \(cm^3\).

\(\therefore\) \(\pi r^2h\) = 49896....(1)

And curved surface area of cylinder is 4752 \(cm^2\).

\(\therefore\) \(2 \pi rh\) = 4752...(2)

On dividing equation (1) by (2), we get

= \(\frac {\pi r^2h}{2\pi rh} = \frac {49896}{4752}\) 

\(\Rightarrow\) \(\frac r2 = 10.5 = \frac {21}2\)

\(\Rightarrow\) r = \( \frac {21}2 \times 2 = 21 \, cm\)

Hence, the radius of the base of cylinder is 21 cm.

Correct option is: C) 21

The height of the cylinderical 

Rod (h) = 11 cm, 

Radius (r) = 7/2 cm 

Volume = πr²h = \(\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\) × 11 

= 423.5 (cm)³

Volume of 50 rods = 50 × 423.5 = 21175 (cm)³

Correct option is: D) 1540 cm³

Given that height of the cylinder is h = 10 cm

and the base radius is r = 7 cm.

\(\therefore\) Volume of cylindrical rod = \(\pi r^2h\)

= \(\frac {22}7\) \(\times\)7 \(\times\) 7 \(\times\) 10

= 1540 \(cm^3\)

Correct option is: D) 1540 cm³

Correct option is: B) 3

Given that h = 4 r

\(\therefore\) Volume of cylindrical iron rod = \(\pi r^2h\) = \(4 \pi r^3\) (\(\because\) h = 4r)

\(\because\) Cylindrical iron rod is melted and recast into n spherical balls.

Then n \(\times\) volume of one spherical ball = Volume of cylindrical iron rod

\(\Rightarrow\) n = \(\frac {Volume \, of \, cylindrical \, iron \, rod}{Volume \, of \, one \, spherical\, ball}\)  = \(\frac {4 \pi r^3}{\frac 43 \pi r^3 }\) (\(\because\) r is also radius of formed spherical ball)

= 3 

Hence, number of balls thus formed is 3.

Correct option is: B) 3

Given dimensions are: 

Cone: 

Radius = r 

Height = h 

Volume (V) = \(\frac{1}{3}\)πr²h 

Cylinder: 

Radius = r 

Height = h 

Volume (V) = πr²h 

Ratio of volumes of cylinder and cone = πr²h : \(\frac{1}{3}\)πr²h 

= 1 : \(\frac{1}{3}\)

= 3 : 1 

Hence, their volumes are in the ratio = 3 : 1.

Correct option is: C) 3 : 1

Let the radii of both cylinder and cone be r

and the heights of both cylinder and cone be h.

\(\therefore\) Volume of cylinder is \(V_1 = \pi r^2h\) 

Volume of cone is \(V_2 = \frac 13 \pi r^2 h\).

Ration of their volumes = \(\frac {V_1}{V_2}\) = \(\frac {\pi r^2h}{\frac 13 \pi r^2h}\) =\(\frac 31\) = 3 : 1

Hence, the ratio of their volumes is 3:1.

Correct option is: C) 3 : 1

Correct option is: C) 2(lb + bh + hl)

The total surface are of cuboid = 2 ( lb + bh+ hl)

Where l = length of the cuboid,

b = breadth of the cuboid and

h = height of the cuboid.

Correct option is: C) 2(lb + bh + hl)

Correct option is: B) 37 cm

Curved surface area of cone is 4070 \(cm^2\)

i.e. \(\pi rl\) = 4070 ....(1)

Also given that diameter of cone is 70 cm.

\(\therefore\) 2r = 70 cm.

= r = \(\frac {70}2 = 35 \, cm\)

\(\therefore\) l = \(\frac {4070}{\pi r} =\frac {4070}{22 \times 35} \) (From (1))

= \(\frac {814}{22}\) = 37 cm.

Hence, the slant height of the cone is 37 cm.

Correct option is: B) 37 cm

C.S.A. of a cone = πrl = 4070 cm² 

Diameter of the cone (d) = 70 cm 

Radius of the cone = r = d/2 = 70/2 = 35 cm 

Let its slant height be ‘l’. 

By problem, 

πrl = 4070 cm²

22/7 × 35 × l = 4070 

110 l = 4070 

l = \(\frac{4070}{110}\) = 37 cm 

∴ Its slant height = 37 cm.

Given,

internal diameter of pipe = 2 cm

internal radius of pipe = \(\frac{2}{2}\) = 1 cm

rate of flow of water = 6 m/s = 600 cm/s

radius of base of cylindrical tank = 60 cm

so,

rise in height in cylindrical tank = \(\frac{rate\,of\,flow\,of\,water\times total\,time\times volume\,of\,pipe}{volume\,of\,cylinderical\,tank}\)

= \(\frac{600\times 30\times 60\times π\times 1\times 1}{π\times 60\times 60}\) = 300 cm = 3 m

The perimeter of the square garden = 48 m

= 4 x side = 48 m

Side = 48/4 m = 12 m

Area of the square garden = side × side

= 12 m × 12 m = 144 m²

Area of the small flower bed = 18 m²

∴ The area of the garden that is not covered by the flower bed = 144 m² – 18 m² = 126 m²

The required fractional part of the garden which is covered by the flower bed 

= Area of the flower bed/ Area of the square garden = 15/144 = 1/8

The ratio of the area covered by the flower bed and the remaining area

= 18/126 = 1/7 i.e.. 1 :7.

Given,

Areas of three faces of cuboid = x, y, z

Let length of cuboid = l, breadth = b,height = h

So,

 = x = l x b

= y = b x h

= z = h x a

Or we can write ,

= xyz = l²b²h² ... ... ... ... . . .(i)

If ‘V’ is volume of cuboid = V = lbh

= V² = l²b²h² = xyz ... ... ... ... ... .from (i)

= V² = xyz Proved

Given,

Let length of cuboid = l cm

Let breadth of cuboid = b cm

Let height of cuboid = h cm

So,

Area of floor = l x b = lb cm²

Product of areas of two adjacent walls = (l x h) x (b x h) = lbh² cm⁴

Product of areas of floor and two adjacent walls = lb x lbh² cm⁶

= l²b²h² = (lbh)² cm⁶

= (volume of cuboid)² Proved

Given,

Dimensions of swimming pool are = 20 m × 15 m ×3 m

Where,

Length = 20 m , Breadth = 15 m , Height = 3 m

Then,

Area of floor & walls of swimming pool = l x b + 2(l x h + b x h)

= 20 x 15 + 2(20 x 3 + 15 x 3) = 300 + 2(60 + 45) = 300 + 210 = 510 m²

So,

Cost of repairing 1 m² area = Rs.25

∴ Cost of repairing 510 m² = 510 x 25 = Rs. 12750

Correct option is: A) 2h(l + b)

Lateral surface area of four walls of a room 

= 2 (l +b) h square units

Correct option is: A) 2h(l + b)

Let a be the side of the square track.

Perimeter of the square track = 4a

Bajinder runs ten times around the square track and covers 4 km.

i.e., 10 × 4a = 4 km

⇒ 40a = 4 × 1000 m [∵ 1 km = 1000 m]

⇒ a = (4 x 100)/40 m = 100 m

∴ Length of the track = 4a

= 4 × 100 m = 400 m