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प्रश्न
The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of the frustum complete the following activity.( \[\pi = \frac{22}{7}\])
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उत्तर
(Circumference)1 = 2πr1 = 132
∴ r1 = `132/(2π) = (132 xx 7)/(2 xx 22)` = 21 cm
(Circumference)1 = 2πr2 = 88
r2 = `88/(2π) = (88 xx 7)/(2 xx 22)`
= 14 cm
Slant height of frustum, `l = sqrt(h^2 + (r_1 - r_2)^2`
= `sqrt((24)^2 + (21 - 14)^2`
= `sqrt((24)^2 + (7)^2`
= `sqrt(576 + 49)` = `sqrt(625)`
= `sqrt(25 xx 25)` = 25 cm
Curved surface area of the frustum = `π(r_1 + r_2)l`
= `22/7 (21 + 14) xx 25`
= `22/7 xx 35 xx 25`
= 2750 cm2
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