The height and diagonal of the axial section of the truncated cone are related as 5:13.

| May 7, 2021 | education

The height and diagonal of the axial section of the truncated cone are related as 5:13. Find the volume of a cone if the areas of its bases are 16n (pi) and 64n (pi) cm ^ 2.

Knowing the areas of the bases of the cone, we determine their radii.

S1 = n * R1^2 = n * BO1^2 = 16 * n cm2.

BO1 ^ 2 = 16.

BO1 = 4 cm.

S2 = n * R2 ^ 2 = n * AO ^ 2 = 64 * n cm2.

AO ^ 2 = 64.

AO = 8 cm.

Let’s draw the height of the HV cone. Let BH = 5 * X cm, then BD = 13 * X cm.

The length of the segment OH = BO1 = 4 cm, then DH = DO + OH = 8 + 4 = 12 cm.

In a right-angled triangle BDH, DH ^ 2 = BD ^ 2 – BH ^ 2 = 169 * X ^ 2 – 25 * X ^ 2 = 144.

144 * X ^ 2 = 144.

X = 1.

Then BH = 5 * 1 = 5 cm.

Determine the volume of the truncated cone.

V = n * BH * (BO12 + BO1 * AO + AO ^ 2) / 3 = n * 5 * (16 + 32 + 64) / 3 = n * 560/3 = n * 186 (2/3) cm3.

Answer: The volume of the cone is equal to n * 186 (2/3) cm3.



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