The radii of the ends of a frustum are in the ratio 5 : 2 and its slant height is 26 cm. Its curved surface area is 4576 cm². Find the LARGER radius. (Take \π = 22/7 .)

Question

Accepted Answer

40 cm. Let R = 5k and r = 2k, so R + r = 7k. CSA = 22/7 × 26 × 7k = 572k. Setting 572k = 4576 gives k = 8, so R = 5 × 8 = 40 cm. 16 cm is the smaller radius r = 2k; 8 cm is k itself, not a radius; 56 cm is R + r = 7k.

The radii of the ends of a frustum are in the ratio 5 : 2 and its slant height is 26 cm. Its curved surface area is $4576 cm^{2}$ . Find the LARGER radius. (Take $π = \frac{22}{7} .$ )

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More Class 10 Frustum of a Cone practice

The radii of the ends of a frustum are in the ratio 5 : 2 and its slant height is 26 cm. Its curved surface area is 4576cm2. Find the LARGER radius. (Take π=722​.)

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 Frustum of a Cone practice

The radii of the ends of a frustum are in the ratio 5 : 2 and its slant height is 26 cm. Its curved surface area is $4576 cm^{2}$ . Find the LARGER radius. (Take $π = \frac{22}{7} .$ )