Example 2: A right triangle, whose perpendicular sides are 30 cm and 40 cm, is made to revolve about - Maths - Surface Areas and Volumes

Question

Example 2:
A right triangle, whose perpendicular sides are 30 cm
and 40 cm, is made to revolve about its hypotenuse Find the surface
area of the double cone so obtained
can any 1 explain me with video so it will be more clear

Monkey · Accepted Answer

Let ABD be the right-angled triangle such that AB = 30 cm and AD = 40 cm Using Pythagoras theorem, we obtain BD2 = AB2 + AD2 BD2 = $(30 {c m)}^{2} + (40 {c m)}^{2} = 900 {c m}^{2} + 1600 {c m}^{2} = 2500 {c m}^{2}$ BD = 50 cm Let OB = x and OA = y

Applying Pythagoras theorem in Î”AOB, we obtain OA2 + OB2 = AB2 x2 + y2 = 302 y2 = 900 âˆ’ x2 â€¦ (1) Applying Pythagoras theorem in Î”AOD, we obtain OA2 + OD2 = AD2 y2 + (50 âˆ’ x) 2 = 402 900 âˆ’ x2 + 2500 + x2 âˆ’ 100x = 1600 {From equation (1), we have y2 = 900 âˆ’ x2} 100x = 1800 x = 18 On putting this value in equation (1), we obtain y2 = 900 âˆ’ x2 = 900 âˆ’ 182 = 900 âˆ’ 324 = 576 y = 24 cm Here, y is the radius of both the cones Now, surface area of the solid = C S A of cone ABC + C S A of cone ADC = Ï€ Ã— 24 Ã— 30 + Ï€ Ã— 24 Ã— 40 = 1680 Ï€ = 1680 Ã— $\frac{22}{7}$ = 5280 cm2 Thus, the surface area of the double cone so obtained is 5280 cm2

Example 2: A right triangle, whose perpendicular sides are 30 cm and 40 cm, is made to revolve about its hypotenuse. Find the surface area of the double cone so obtained. can any 1 explain me with video..so it will be more clear..

Example 2:

A right triangle, whose perpendicular sides are 30 cm and 40 cm, is made to revolve about its hypotenuse. Find the surface area of the double cone so obtained.

can any 1 explain me with video..so it will be more clear..