Question

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Solve question no. 30 !

Need Both Question Which are in Choice ✔

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Answer 1

Question No. 30)

Refer to the attachment for the figure.

Answer :

Circumference of the base of tent = 44 metres

We know that,

Circumference = 2 π r

Thus,

2 π r = 44

2 × 22/7 × r = 44

22/7 × r = 22

r = 22 × 7/22

r = 7

Radius of base of tent = 7 m.

Now,

Height of conical tent is 10 metres.

Slant height = L = √(h^2 + r^2)

L = √(100 + 49)

L = √149

L = 12.2 metres ( approx. )

Now,

Surface area of tent = Area of rectangular canvas

π r l = L × B

22/7 × 7 × 12.2 = L × 2m

22 × 6.1 = L

L = 134.2 metres

Hence,

The length of the canvas is 134.2 metres.

_________________________

Answer 2

${\huge {\mathfrak {Answer :-}}}$

Let 'R' be the external radius and ' r ' the internal radius of the hollow wooden cylindrical part of the pencil whose interior is filled with graphite.

➡️ Then, R = 7/2 mm = 7/20 cm.

➡️ And, r = 1/2 mm = 1/20 cm.

Let, 'h' be the height of the hollow wooden cylinder.

➡️ Then, h = 14cm.

Volume of the wood

= $\pi$ (R^2 - r^2) h

= 22/7 * [(7/20)^2 - (1/20)^2] * 14 Cu.cm

= 22/7 * [(49 - 1)/400] * 14 Cu.cm

= 22/7 * 48/400 * 14 Cu.cm

= 5.28 Cu.cm

➡️ So, Volume of the wood is 5.28 Cu.cm .

➡️ Now, we have,

Volume of the graphite

= $\pi$ * r^2 * 14 Cu.cm

= 22/7 * 1/20 * 1/20 * 14 Cu.cm

= 0.11 Cu.cm

➡️ $<b>Hence, the volume of the wood is 5.28 cu.cm and that of the graphite is 0.11 cu.cm </b>$ ⬅️

$<i>$ That's it..

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