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प्रश्न
A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm, and height 3 cm. Find the area of its whole surface and volume.
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उत्तर
The radii of the bottom and top circles are r1 = 10 cm and r2 = 6 cm respectively. The height of the frustum cone is h= 3 cm. Therefore, the volume of the bucket is
`V=1/3pi(r_1^2+r_1r_2+r_2^2)xxh`
`=1/3pi(10^2+10xx6+6^2)xx3`
= 616 cm3
Hence Volume = 616 cm3
The slant height of the bucket is
`l=sqrt((r_1-r_2)+h^2)`
`=sqrt((10-6)^2+3^2)`
`=sqrt(25)`
= 5cm
The total surface area of the frustum cone is
`pi(r_1+r_2)xxl+pir_1_pir_2^2`
`=22/7xx(10+6)xx5+22/7xx10^2+22/7xx6^2`
`=4752/7`Square cm
= 678.85 Square cm
Hence Total surface area = 678.85
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