The height of the truncated cone is 4 cm. the radius of one base of the cone is 2 times the radius of the other

| July 1, 2021 | education

The height of the truncated cone is 4 cm. the radius of one base of the cone is 2 times the radius of the other, and the sum of the areas of the bases is equal to the area of the lateral surface. find the radius of the base of this truncated cone.

Let the radius of the smaller base be X cm, BO1 = X cm, then the radius of the larger base is OA = 2 * X cm.

Determine the areas of the bases.

S1 = n * BO1 ^ 2 = n * X ^ 2.

S2 = n * AO ^ 2 = n * 4 * X ^ 2.

Then S1 + S2 = n * 5 * X ^ 2.

Let’s draw the height of the HВ. Then AH = AO = OH = 2 * X – X = X cm.

In a right-angled triangle ABН, AB ^ 2 = BH ^ 2 + AH ^ 2 = 16 + X ^ 2. (1)

Let us determine the area of ​​the lateral surface of the cone.

Side = n * (BO1 + AO) * AB = n * 3 * X * AB.

Since S1 + S2 = S side, then (n * 5 * X ^ 2 = n * 3 * X * AB).

5 * X = 3 * AB. Squaring both sides and substitute equation 1.

25 * X ^ 2 = 9 * AB ^ 2.

25 * X ^ 2 = 9 * (16 + X ^ 2) = 144 + 9 * X ^ 2.

16 * X ^ 2 = 144.

X ^ 2 = 144/16 = 9.

X = BO1 = 3 cm.

AO = 2 * 3 = 6 cm.

Answer: The radii of the bases are 3 cm and 6 cm.



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