Question

from a solid cube of side 7 cm a conical cavity of height 7 cm and radius 3 cm is hollowed out find the surface area of remaining part of the solid​

Answer 1

Answer:

277cm^3

volume of cube =(7)^3=343cm^3

volume of cone=1/3*22/7*3*3*7=66cm^3

Remaining solid=343-66=277cm^3.....

Answer 2

$\large{\underline{\rm{\purple{\bf{Given:-}}}}}$

Radius, r = 3 cm

Height, h = 7 cm

$\large{\underline{\rm{\purple{\bf{To \: Find:-}}}}}$

The surface area of remaining part of the solid​.

$\large{\underline{\rm{\purple{\bf{Analysis:-}}}}}$

Volume of remaining solid = volume of cube – volume of cone

$\large{\underline{\rm{\purple{\bf{Solution:-}}}}}$

We know that,

Volume of remaining solid = volume of cube – volume of cone

For Cube

Side, a = 7 cm

We know that,

Volume of cube = $\sf a^{3}$, where a = side of cube

Volume of cube = $\sf (7)^{3} = 343 \: cm^{3}$

For cone

Radius, r = 3 cm

Height, h = 7 cm

Volume of cone = $\sf \dfrac{1}{3} \pi r^{2} h$

$\implies \sf \dfrac{1}{3} \pi (3)^{2}7$

$\implies \sf 3 \times \dfrac{22}{7} \times 7 =66 \: cm^{3}$

Volume of remaining solid = volume of cube – volume of cone

$\implies \sf 343-66$

$\implies \sf 277 \: cm^{3}$

The surface area of remaining part of the solid​ is 277 cm³