The upper part of a metal funnel is a frustum with rim radii 11 cm and 3 cm and a vertical depth of 15 cm. Find the area of sloping sheet in it, in cm². (Take \π = 22/7 .)

Question

Accepted Answer

748. l = √15^2 + 8^2 = √289 = 17 cm. CSA = 22/7 × 17 × (11 + 3) = 22/7 × 17 × 14 = 748 cm². 660 uses the depth 15 cm as the slant height; 427.43 uses (R - r) = 8 instead of (R + r) = 14; 1496 doubles the correct answer.

The upper part of a metal funnel is a frustum with rim radii 11 cm and 3 cm and a vertical depth of 15 cm. Find the area of sloping sheet in it, in $cm^{2}$ . (Take $π = \frac{22}{7} .$ )

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More Class 10 Frustum of a Cone practice

The upper part of a metal funnel is a frustum with rim radii 11 cm and 3 cm and a vertical depth of 15 cm. Find the area of sloping sheet in it, in cm2. (Take π=722​.)

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 Frustum of a Cone practice

The upper part of a metal funnel is a frustum with rim radii 11 cm and 3 cm and a vertical depth of 15 cm. Find the area of sloping sheet in it, in $cm^{2}$ . (Take $π = \frac{22}{7} .$ )