The radii of the bases of the cut cone are 2 cm and 7 cm, the diagonal of the axial section is 15 cm.

The radii of the bases of the cut cone are 2 cm and 7 cm, the diagonal of the axial section is 15 cm. Find the area of the lateral surface.

The lateral surface area of ​​the truncated cone is calculated by the formula:

Sb.p. = π (r1 + r2) h.

To do this, you need to calculate the length of the CM height.

For this, consider the triangles ΔАОМ and ΔКОС. These triangles are similar, since the angles KOS and AOM are equal to each other, as are the vertical angles, and the angles OKC and ОМА are right.

Thus, using the similarity coefficient, we can find the length of the AO and OC segments.

k = AM / KC;

k = 7/2 = 3.5.

Since the AC diagonal is 15 cm, then express it like this:

x is the length of the CO segment;

3.5x – length of the AO segment;

x + 3.5x = 15;

4.5x = 15;

x = 15 / 4.5 ≈ 3.33;

CO = 3.3 cm;

AO = 3.3 3.5 = 11.7 cm.

Now, using the Pythagorean theorem, we find the length of the segments KO and OM, the sum of which is the length of the CM height:

OC ^ 2 = KO ^ 2 + KC ^ 2;

KO ^ 2 = OS ^ 2 – KC ^ 2;

KO ^ 2 = 3.3 ^ 2 – 2 ^ 2 = 10.89 – 4 = 6.89;

KO = √6.89 ≈ 2.62 cm;

AO ^ 2 = AM ^ 2 + OM ^ 2;

OM ^ 2 = AO ^ 2 – AM ^ 2;

OM ^ 2 = 11.7 ^ 2 – 7 ^ 2 = 136.89 – 49 = 87.89;

ОМ = √87.89 ≈ 9.38 cm.

KM = KO + OM;

KM = 2.62 + 9.38 = 12 cm.

Now let’s find the area of ​​the lateral surface:

Sb.p. = 3.14 * (2 + 7) * 12 = 3.14 * 9 * 12 = 339.12 cm2.

Answer: the area of ​​the truncated cone is 339.12 cm2.