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प्रश्न
From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.
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उत्तर
We have,
the height of the cone = the height of the cylinder = h =2.8 cm and the radius of the base, r = `4.2/2 = 2.1 "cm"`
The slant height of the cone, `l = sqrt(r^2 + h^2)`
`=sqrt(2.1^2 + 2.8^2)`
`= sqrt(4.41 +7.84)`
`=sqrt(12.25)`
= 3.5 cm
Now, the total surface area of the remaining solid = CSA of cylinder + CSA of cone + Area of base
`= 2pirh + pirl + pir^2`
`=pir (2h +l +r)`
`= 22/7 xx2.1xx(2xx2.8xx3.5+2.1)`
`= 22xx0.3xx(5.6+5.6)`
`=6.6xx 11.2`
`= 73.92 "cm"^2`
So, the total surface area of the remaining solid is 73.92 cm2.
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Students found the shapes of the objects very interesting and they could easily relate them with mathematical shapes viz sphere, hemisphere, cylinder etc. |
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