Question

please give me answers 6 to 11.​

Answer 1

Answer:

11. ✌️answer 281.6✌️

csa of pillar =2 pie rh

2*22/7*0.35*4 =8.8 m2

cost of cementing =8.8*32

=281.6rs

I have the solution but it is not coming more than 5 so sorry I have typed it

Hope it is helpful ♥️

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Answer 2

$\huge \sf{ \underline{ \underline{Question 6.}}}$

the diameter of a roller is 77 cm and its length is 105 cm. it takes 600 revolutions moving once over, to level a playground. find the area of the playground in m².

Given:-

• the length of the roller is 105m
• diameter is 77cm

$\sf \: \therefore \: radius \: of \: the \: roller \: is \frac{77}{2}cm$

let height be it's length

$\sf \: CSA \: of \: the \: roller=2πrh$

$\sf \: =2× \frac{22}{7}× \frac{77}{2}×105$

$\sf = 2 \times \frac{ \cancel{22} \: \: \: ^{11} }{ \cancel7} \times \frac{ \cancel{77} \: ^{11} }{ \cancel2} \times 105$

$=2×11×11×105$

$=22×1155$

$=25410cm²$

$\sf \: in \: metres= \frac{25410}{100}=254.1m²$

takes 600 revolutions to level the playground then the area of the playground is =254.1×600

=152460m²

hence the final answer is 152460m².

Question 7

Find the area of the playground in m ind the volume of the earth dug out from a circular well of 24.5 m deep and 3 m radius.

$\huge \sf{ \underline{ \underline{Solution:- }}}$

★given the radius of the well is 3m and height (depth) of the well is 24.5m

volume of the well=πr²h

$\sf = \frac{22}{7} \times 3×3×24.5$

$= \frac{4851}{7}=693m³$

hence the volume of the earth dug out is 693m³

Question 8

The area of the base of a cylindrical water tank is 9.63 m and its height is 4 m. How many

kilolitres of water can it hold?

$\huge \sf{ \underline{ \underline{Solution:- }}}$

Volume of cylinder. tank = πr 2h

= 9.63x4=38.52m³

=38.52x1000 =38520litre

=385.2 kilolitre

Question 9

The inner radius of a cylindrical tunnel is 9 m and it is 21 m long. Find the cost of painting the

inner surface of the tunnel at the rate of 8 per m'

Solution:-

Given :

The inner radius of a cylindrical tunnel is 9 m and 21 m long.

To find :

The cost of painting the inner surface of the tunnel at the rate of rupees 8 per meter square ?

Solution :

The painting will be done only on the curve

surface portion of the tunnel.

So, The area of the painting is

$A=2\pi \: rh$

r= 9 m is the radius

h= 21 m i the height

$\sf = 2 \times \frac{22}{7} \times 9 \times 21$

$= 1188 {m}^{2}$

The rate of rupees 8 per meter square,

Cost of the painting is:-

$c = rate \times area$

$= 8 \times 1128 = 9024$

Step-by-step explanation:

Sorry I don't know about the answers of 10,11