Math, asked by shivamtyagidestiny20, 4 months ago

a right circular cone has height of 6cm if two cut parallal to base​

Answers

Answered by Naisha28
0
Answer:
8:117:91

Explanation :

Main Cone = Cone1 = ABC1

Height = h6 = 6 cm

2 right angled triangles are ACB and ACC1 and their bases are B = B.

Main cone has cut at 1 cm, so the new cone within main cone is at a height of 1 cm from the base of Main cone ABC1.

The new cone =Cone2= ADE1 whose height is
h5 = h6 - 1 cm = 6 - 1 = 5 cm
h5 = 5 cm

Main cone has 2nd cut at 4 cm from the main base, so the another new cone within main cone is at a height of 4 cm from the base of Main cone ABC1.

The 2nd new cone = Cone3 = AFG1 whoes height is

h2 = h6 - 4 cm = 6 - 4 = 2 cm

h2 = 2 cm

In the right angled triangles ACB, ACD and AGF we have to find out the value of line CB=B, ED=B1 and GF= B2 respectively.

Line CB = Base B

Line ED = Base B1

Line GF = Base B2

AC/ CB = tan 45° => CB = AC / tan 45°

Line CB = Base B = 6 / 1 (i.e tan 45°=1)

Base B = 6 cm

Similarly,

Base B1 = AE/ED = tan 45° => ED = AE/ tan 45°

Line AE = 6 - 1 = 5 cm

Base B1 = line ED = 5/1 = 5 cm

Base B2 = AG/GF = tan 45°
=> GF = AG/ tan45°
Line AG = 6 - 4 = 2 cm

Base B1 = line GF = 2/1 = 2 cm

Volume of Cone1 = 1/3 * π * B² * h6

= π/3*6²*6 = (π/3)*216 cm³

Volume of Cone2 = 1/3 * π * B1² * h5

= (π/3)*(5)²*(5)= (π/3)*(125) cm³

Volume of Cone3 = 1/3 * π * B2² * h2

= (π/3)*(2)²*(2)= (π/3)*(8) cm³

due to 2 parallel cuts at 1 cm and 4 cm from base B there are 3 sections of main cone from top to bottom. These are

section 1 = cone AFG1 = Volume of Cone 3

section 2 = cone ADE1 = Volume of Cone 2 - Volume of Cone 3

= π/3*125 - π/3*8 = π/3*(125 - 8) =π/3*117 cm³

section 3 = Vol of Main Cone - Vol.of Cone 2

= π/3*216 - π/3*125 = π/3*(216 - 125)

= π/3*91 cm³

section 3 = π/3*91 cm³

Ratio among the sections

section-1 : section-2 : section-3

(π/3)*(8) : π/3*117 : π/3*91

Divide all ratio by π/3 => 8 : 117 : 91




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