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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.

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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.( \[\pi = \frac{22}{7}\]) 

योग
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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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अध्याय 7: Mensuration - Practice set 7.2 [पृष्ठ १४८]

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बालभारती Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
अध्याय 7 Mensuration
Practice set 7.2 | Q 3 | पृष्ठ १४८

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