Question

plz solve this maths question
I will be thankful to u if u solve it ​

Answer 1

Solution:-

first hemisphere length=π×0.5=0.5πcm

second hemisphere=π×1=πcm

third hemisphere length=π×1.5=1.5πcm

fourth hemisphere length=π×2=2πcm

here we are seeing that AP is fromed of common difference 0.5π

length of.spiral=sum of 15terms of this AP

length of spiral=15/2(2×0.5π+14×0.5π)cm

=15×4π=60πcm

{sorry for mistake}

Answer 2

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Solution :-

Let L₁, L₂, L₃, L₄, ........, L₁₃ be the lengths of semicircles of radii 0.5 cm, 1 cm, 2 cm .... and 13/2 cm respectively.

Then, we have

L₁ = π × 0.5 = π/2 cm,

L₂ = π × 1 = 2(π/2) cm,

L₃ = π × 1.5 = 3(π/2) cm,

L₄ =  π × 2 = 4(π/2) cm,

_________________

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L₁₃ =  π × 13/2 = 13(π/2) cm,

Therefore,

Total length of the spiral

= L₁ + L₂ + L₃ + L₄ + ....., + L₁₃

= [π/2 + 2(π/2) + 3(π/2) + 4(π/2) + ......., + 13(π/2)] cm

= π/2(1 + 2 + 3 + 4 + ........ + 13) cm

= π/2 × 13/2 × (1 + 13) cm

= (1/2 × 22/7 × 13/2 × 14) cm

= 143 cm.

Hence, the required length of the spiral is 143 cm.

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