A right triangle having sides 15 …

Ans.

Consider the following right angled triangle ABC is rotated through its hypotenuse AC

BD {tex}\perp{/tex} AC. In this case BD is the radius of the double cone generated.

Using Pythagoras theorem for {tex}\triangle ABC{/tex} 

=> AC² = AB² + BC

=>  AC² = 225 + 400 = 625

 => AC = 25 cm

Let AD = x cm CD = (25 – x) cm

Using Pythagoras theorem in {tex}\triangle ABD{/tex}

=> AD² +BD² = AB²

=> x² + BD²= 225

=> BD²= 225 – x² …………(1)

Using Pythagoras theorem for {tex}\triangle CBD{/tex}

=> BD² + CD² = BC²

=> BD² +(25-x)²= 400

=> BD² = 400 - (25-x)² …………(2)

From (1) & (2)

=> 225 - x²= 400 - 625 - x² + 50x

=> 50x = 450

=> x = 9 cm

AD = 9, CD = 16

From (1) BD = 12

Surface area of the double cone formed = L.S.A of upper cone + L.S.A of the lower cone

{tex} => \pi \times BD \times AB + \pi \times BD \times BC{/tex}

{tex} => \pi \times BD \times (AB + BC){/tex}

{tex}=> \pi \times 12(20+15){/tex}

{tex}=> \pi \times 12\times 35{/tex}

{tex}=>420 \pi {/tex}cm²