In a right angle, the sides conrtaining right angles are 5 cm and 12 cm. It is rotated aboutits hypotenuse taking it as axis. Find the total surface area and volume of formed figure.

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Let `Delta ABC` is given in which
AB = 5cm
BC = 12cm
and `angleABC= 90^(@)`
Now, from Phythagoarstheorem
`AC^(2)=AB^(2)+BC^(2)`
`=5^(2)+12^(2)`
= 25 + 144 = 169
`rArr AC=13cm`
Let BO = r
Now, area of `Delta ABC =(1)/(2)xxABxxBX=(1)/(2)xx5xx12`
Again, area of `Delta ABC = (1)/(2)xxACxxBO=(1)/(2)xx13xx r`
`therefore (1)/(2)xx13xx r=(1)/(2)xx5xx12`
`rArr r = (60)/(13) cm`
Now, the volume of solid = volume of cone ABD + volume of cone CBD
`=(1)/(3)pi r^(2)xxAO+(1)/(3)pi r^(2)xxOC=(1)/(3)pi r^(2)(AO+OC)`
`=(1)/(3)pi r^(2)xxAC=(1)/(3)xx pi xx(60)/(13)xx(60)/(13)xx13=(1200pi)/(13)cm^(3)`
and total surface area = curved surface of cone ABD + curved surface of cone CBD
`= pi r(AB)+ pi r(BC)`
`= pi r(AB+BC)`
`= pi xx(60)/(13)(5+12)`
`=(1020pi)/(13)cm^(2)`