In a truncated cone, the diagonal of the axial section is 10 cm, the radius of the smaller base is 3 cm

| June 16, 2021 | education

In a truncated cone, the diagonal of the axial section is 10 cm, the radius of the smaller base is 3 cm, and the height is 6 cm. find the full surface of the frustum.

The axial section of the truncated cone is an isosceles trapezoid ABCD.

Let’s draw the height BH of the trapezoid. In a right-angled triangle ВDН, according to the Pythagorean theorem, we determine the length of the leg DH.

DH ^ 2 = BD ^ 2 – BH ^ 2 = 100 – 36 = 64.

DН = 8 cm. The segment ОН = BO1 = 3 cm, then ОD = ОА = DH – ОН = 8 – 3 = 5 cm.

Segment AH = OA – OH = 5 – 3 = 2 cm.

In a right-angled triangle ABH, we determine the length of the hypotenuse AB.

AB ^ 2 = BH ^ 2 + AH ^ 2 = 36 + 4 = 40.

AB = 2 * √10 cm.

Determine the area of ​​the larger base. Sosn1 = n * OA ^ 2 = n * 25 cm2.

Determine the area of ​​the smaller base. Sosn2 = n * OB1 ^ 2 = n * 9 cm2.

Determine the lateral surface area: S side = n * AB * (OA + OB1) = n * 2 * √10 * 8 = n * 16 * √10 cm2.

Determine the total area of ​​the truncated cone: S = Sbok + Sbn1 + Sbn2 = n * 16 * √10 + 25 + 9 = 2 * n * (8 * √10 + 17) cm2.

Answer: The total area is 2 * n * (8 * √10 + 17) cm2.



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