Question

Q30. A right triangle, whose sides are 3cm and 4cm (other than hypotenuse) is made to revolved about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate).​

Answer 1

$\large\underline{\sf{Solution-}}$

Let us consider a right-angle triangle ABC right-angled at B such that AB = 3 cm and BC = 4 cm.

So,

In right triangle ABC,

By Pythagoras Theorem,

$\rm :\longmapsto\: {AC}^{2} = {AB}^{2} + {BC}^{2}$

$\rm :\longmapsto\: {AC}^{2} = {3}^{2} + {4}^{2}$

$\rm :\longmapsto\: {AC}^{2} = 9 + 16$

$\rm :\longmapsto\: {AC}^{2} = 25$

$\rm :\longmapsto\: {AC} = 5 \: cm$

Now,

$\rm :\longmapsto\:Area_{(\triangle ABC)} = \dfrac{1}{2} \times AB \times BC = \dfrac{1}{2} \times AC \times r$

$\rm :\longmapsto \: 4 \times 3 = 5 \times r$

$\rm :\longmapsto\:r = \dfrac{12}{5} \: cm$

Now,

Since, Triangle is revolved along AC so that double cone is formed having radius BD (r) and height AD and other cone having radius BD (r) and height CD.

So,

Volume of double cone thus generated is

$\sf \: Volume_{(double \: cone)} = Volume_{(cone1)} + Volume_{(cone2)}$

$\rm \: Volume_{(double \: cone)} = \dfrac{1}{3} \pi \: {(r)}^{2}AD + \dfrac{1}{3} \pi \: {(r)}^{2}CD$

$\rm \: Volume_{(double \: cone)} = \dfrac{1}{3} \pi \: {(r)}^{2}(AD + CD)$

$\rm \: Volume_{(double \: cone)} = \dfrac{1}{3} \pi \: {(r)}^{2}(AC)$

$\rm \: Volume_{(double \: cone)} = \dfrac{1}{3} \times3.14 \times {(2.4)}^{2} \times 5$

$\rm \: Volume_{(double \: cone)} = 30.14 \: {cm}^{2}$

CALCULATION of Surface Area

Dimension of first cone :-

• Radius of first cone, AD = 2.4 cm

• Slant height of cone, AB = 3 cm

Dimensions of second cone :-

• Radius of first cone, AD = 2.4 cm

• Slant height of cone, BC = 4 cm

So,

Surface Area of double cone is

$\rm \: Surface Area_{(doublecone)} = Surface Area_{(cone1)} + Surface Area_{(cone2)}$

$\rm \: Surface Area_{(doublecone)} = \pi \: AD \times AB + \pi \: AD \times BC$

$\rm \: Surface Area_{(doublecone)} = \pi \: AD \times (AB + BC)$

$\rm \: Surface Area_{(doublecone)} = \pi \: AD \times (3 + 4)$

$\rm \: Surface Area_{(doublecone)} = 3.14 \times 2.4 \times 7$

$\rm \: Surface Area_{(doublecone)} = 52.75 \: {cm}^{2}$

Additional Information :-

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²