Answer:
The volume and the curved surface area of the solid formed is [tex]301.71\ \text{cm}^3[/tex] and [tex]188.57\ \text{cm}^2[/tex] respectively.
Step-by-step explanation:
Given that,
A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm.
As the triangle is revolved at 8 cm it forms a cone of height 8 cm and radius 6cm.
Height, h = 8 cm
Radius, r = 6 cm
The volume of the solid formed is given by :
[tex]V=\dfrac{1}{3}\pi r^2 h\\\\=\dfrac{1}{3}\times \dfrac{22}{7}\times 8^2\times 8\\\\=301.71\ \text{cm}^3[/tex]
The curved surface of the solid formed is :
[tex]A=\pi rl[/tex]
l is slant height
[tex]A=\pi r\times \sqrt{h^2+r^2} \\\\A=\dfrac{22}{7}\times 6\times \sqrt{8^2+6^2} \\\\=188.57\ \text{cm}^2[/tex]
Hence, the volume and the curved surface area of the solid formed is [tex]301.71\ \text{cm}^3[/tex] and [tex]188.57\ \text{cm}^2[/tex] respectively.
