Question

Ram and mehmood are 2 neighbours. Ram wears a cap which is in the shape of hemisphere of radius 7cm. And mehmood wears a cap which is in the shape of a frustum of a cone. Lf radius of its open side is 10cm and radius of the upper base is 4cm and its sland height is 15cm.Find the curved surface area of both caps

Answer 1

CSA of Rams cap=2πr×r. =2×22/7×7×7. =154cm2//. CSA of frustum of a cone=π×(R+r)l. =22/7×(10+4)15. =330cm2//.

Answer 2

The curved surface of the Ram's cap(hemisphere) = 308 cm²

The curved surface of the Mehmood's cap (frustum)= 660 cm²

Step-by-step explanation:

Ram wear a cap in the shape of hemisphere

radius of hemisphere = 7 cm

curved surface area of the hemisphere cap = $2\pi r^2$

= $2\times\frac{22}{7} \times7\times 7$

= 22×2×7

= 22×14

= 308 cm²

Mehmood wear a cap in the shape of frustum

the radius of open side of the cap(R) = 10 cm

radius of the upper base (r) = 4 cm

slant height of the frustum cap (l) = 15 cm

curved surface area of the frustum cap = $\pi (R+r)l$

putting the values

= $\frac{22}{7}\times(10+4)\times15$

= 22×2×15

= 660 cm²

hence,

The curved surface of the Ram's cap(hemisphere) = 308 cm²

The curved surface of the Mehmood's cap (frustum)= 660 cm²

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