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The perimeter of the regular pentagon = 1540 cm

⇒ 5 × side = 1540 cm

⇒ side = 1540/5 cm = 308 cm

∴ Each side of the regular pentagon is 308 cm.

The perimeter of the lawn in front of Molly’s house

= 2(length + breadth)

= 2(12 + 8) m = 2(20) m = 40 m

The perimeter of the lawn in front of Dolly’s house = 2(length + breadth)

= 2(15 + 5) m = 2 × 20 m = 40 m

∴ Total length of fencing for both the lawns = (40 + 40) m = 80 m

80 m fencing is required for both.

Given: For the cylindrical tablets, 

radius (r) = 7 mm, 

thickness = height(h) = 5 mm 

For the cylindrical wrapper, 

diameter (D) = 14 mm, height (H) = 10 cm 

To find: Number of tablets that can be wrapped.  

Radius of wrapper (R) = Diameter/2 

= 14/2 = 7 mm 

Height of wrapper (H) = 10 cm 

= 10 × 10 mm 

= 100 mm 

Volume of a cylindrical wrapper = πR²H 

= π(7)² × 100 

= 4900π mm³

Volume of a cylindrical tablet = πr² h 

= π(7)² × 5 

= 245 π mm³ 

(No. of tablets that can be wrapped) = (voume of a cylindrical wrapper)/ (valume of a cylindrical tablet) 

= 4900π)/245π

= 20 

∴ 20 tables can be wrapped in the wrapper

Correct option is: B) \( 97 \frac 37 \, cm^2\)

Inner radius of hemisphere is r = 2 cm and outer radius of hemisphere is R = 3 cm.

\(\therefore\) Total surface area of hollow hemispherical cell 

= \(2 \pi r^2 + 2 \pi R^2 + \pi (R^2-r^2)\)

= \(\pi (2 r^2 + 2 R^2 + R^2-r^2)\) 

= \(\pi (3 R^2 + r^2)\)

= \(\frac {22}7\) ( \(3 \times 3^2 + 2^2\) )

= \(\frac {22}7\) (27 + 4)

= \(\frac {22\times 31}7 = \frac {681}7 = 97 \frac 37 \, cm^2\)

\(\therefore\) TSA of hollow hemispherical cell is \( 97 \frac 37 \, cm^2\)

Correct option is: B) 97 \(\frac{3}{7}\)

Let base x , altitude p and area A
Area=1/2(x*p)=60
p = 120/x
Now applying pythagoras theorem 
(x/2)²+(120/x)²=13²
Solving equation we get:

x=24, x=10

i) Volume: 3-d shape 

ii) Area: L.S.A. / T.S.A. 

iii) Volume: 3-d shape 

iv) Volume: 3-d shape 

v) Volume: 3-d shape

Given,

Length of cylinder = 21 dm = 210 cm

Outer diameter = 10 cm

Outer radius = R = \(\frac{10}{2}\) = 5 cm

Inner diameter = 6 cm

Inner radius = r = \(\frac{6}{2}\) = 3 cm

∴ Area of base = π(R² - r²) = π(25 - 9) = 16π cm²

Volume of cylinder = area of base x height = 16π x 210 = 16 x \(\frac{22}{7}\) x 210 = 10560 cm³

A hollow cylinder is cut along its height to form a rectangular sheet.

Area of cylinder = Area of rectangular sheet

4224 cm² = 33 cm × Length

Lenght = 4224 cm²/33 cm = 128cm

Thus, the length of the rectangular sheet is 128 cm. Perimeter of the rectangular sheet = 2 (Length + Width)

= [2 (128 + 33)] cm

= (2 × 161) cm

= 322 cm

Correct option is: C) \(\frac{2}{3}\) cm

Volume of cylinder = volume of the sphere.

= \(\pi r^2h = \frac 43 \pi r^3\) 

= \(4 \times 4 \times h = \frac 43 \times 2 \times 2 \times 2\).

= h = \(\frac {16}3 \times \frac2{16} = \frac 23\)

Hence, the height of the cylinder is \(\frac 23\)cm

Correct option is: C) \(\frac{2}{3}\) cm

Correct option is: (C) Pythagoras

To find out the slant height of a cone, we use Pythagoras theorem.

Correct option is: (C) Pythagoras

Correct option is: (C) 3 : 1

Let r & h be the radius and height of cone and cylinder.

Now, \(\frac {volume\, of\, cylinder}{volume\, of\, cone} = \frac {\pi r^2h}{\frac 13 \pi r^2h} = \frac 31\)

Hence, the ratio of volumes of cylinder and cone is 3 : 1.

Correct option is: (C) 3 : 1

Correct option is: D) 1 : 3

\(\frac {Volume \, of \, cone}{Volume \, of\, cylinder}\) = \(\frac {\frac 13 \pi r^2h}{\pi r^2h} = \frac 13 = 1:3\)

Hence, required ratio is 1; 3

Correct option is: D) 1 : 3

Correct option is: C) 3 : 1 : 2

Let 2r be the diameter of cylinder, cone of sphere.

\(\therefore\) Height of sphere = 2r

Then height of cone = Height of cylinder = 2r.

Now, \(V_1 : V_2 : V_3\) = Volume of cylinder : Volume of cone : Volume of sphere

= \(\pi r^2h : \frac 13 \pi r^2 h : \frac 43 \pi r^3\) (\(\because\) h = 2r)

= \(2 \pi r^3 : \frac {2 \pi r^3 }{3} = \frac {4 \pi r^3 }{3}\)

= 2 : \(\frac 23 : \frac 43\) (On dividing by \(\pi r^3 \))

= 6 : 2 : 4 (On multiplying by 3)

= 3 : 1 : 2 (On dividing by 2)

Hence, the ratio of their volumes is 3 : 1 : 2.

Correct option is: C) 3 : 1 : 2

We have given, perimeter of the rectangular field 

= perimeter of the square field

⇒ 2(length + breadth) = 4 × side

⇒ 2(8 + 2) m = 4 × side

⇒ 2(10) m = 4 × side ⇒ 20 m = 4 × side

⇒ 20/4 m = side ⇒ side = 5 m

Now, area of the rectangular field

= length × breadth = 8m × 2m = 16m²

Area of the square field= side × side = 5m × 5m = 25m²

∴ The square field has greater area than that of rectangular field.

Given,

Depth of well = 20 m

Diameter of well = 7 m

Radius of well = \(\frac{7}{2}\) m

Dimension of rectangular field = 22 m × 14 m

so,

Amount of earth dug out from well = πr²h = \(\frac{22}{7}\) x \(\frac{7}{2}\) x \(\frac{7}{2}\) x 20 = 770 m³

when this earth is spread on rectangular field:

Then height of platform formed on rectangular field = \(\frac{volume\,of\, earth\,dug\,out}{area\,of\,rectangular\,out}\) = \(\frac{770}{22\times 14}\) = 2.5 m

Given,

Dimensions of class room = 7 m × 6 m ×3.5 m

Where,

Length = 7 m, Breadth = 6 m , Height = 3.5 m

Area of four walls (including doors & windows) = 2(l x h + b x h)

= 2(7 x 3.5 + 6 x 3.5) = 91 m²

Then,

Area of walls without doors & windows = area including doors & windows – area occupied by doors & windows

Area of only walls = 91 – 17 = 74 m²

So,

Cost of white washing 1 m² area of walls = Rs.1.50

∴ Total cost of white washing = 74 × 1.50 = Rs.111

(i) Let initially the edge of the cube be l. Initial surface area = 6l²

If each edge of the cube is doubled, then it becomes 2l.

New surface area = 6(2l)² = 24l² = 4 × 6l²

Clearly, the surface area will be increased by 4 times.

(ii) Initial volume of the cube = l³

When each edge of the cube is doubled, it becomes 2l.

New volume = (2l)³ = 8l³ = 8 × l³

Clearly, the volume of the cube will be increased by 8 times.