Question

A right triangle, whose base and height are 15cm and 20cm respectively is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed​

Answer 1

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⎟⎟ ✰✰ ＡＮＳＷＥＲ ✰✰ ⎟⎟

☘ $\tt{Let\:ABC\:be\:right \:angled \:triangle}$

☪ $\tt{AB\:=\:15cm}$

❥ $\tt{AC\:=\:20cm}$

$\underline\bold\purple{Using:Thales\:theorem\:in\:\triangle\:ABC}$

$:\implies\:\tt{BC^2\:=\:AB^2\:+\:AC^2}$

$:\implies\:\tt{BC^2\:=\:15^2\:+20^2}$

$:\implies\:\tt{BC^2\:=\:225\:+\:400}$

$:\implies\:\tt{BC^\:=\:625}$

$:\implies\:\tt{BC\:=\:\sqrt{625}}$

$:\implies\:\tt{BC\:=\:25cm.}$

$\tt{Let,}$

• $\huge\tt{OA\:=\:x}$
• $\huge\tt{OB\:=\:y}$

In triangles ABO and ABC, we have ∠BOA = ∠BAC and ∠ABO = ∠ABC

So, by angle - angle - criterion of similarity, we have ∆BOA ~ ∆BAC

⛬ $\huge\tt{\:\:\frac{BO}{BA}\:=\:\frac{OA}{AC}\:=\:\frac{BA}{BC}}$

$:\implies\:\tt{\frac{y}{15}\:=\:\frac{x}{20}\:=\:\frac{15}{25}}$

$:\implies\:\tt{\frac{y}{15}\:=\:\frac{x}{20}\:=\:\frac{3}{5}}$

$:\implies\:\tt{\frac{y}{15}\:=\:\frac{3}{5}\:and\:\frac{x}{20}\:=\:\frac{3}{5}}$

$:\implies\:\tt{y\:=\:\frac{3}{5}\:×\:15\:and\:x\:=\:\frac{3}{5}\:×\:20}$

$:\implies\:\tt{y\:=\:9\:and\:x\:=\:12}$

Thus, we have

$\:\:\:\:$OA = 12cm and OB = 9cm. Also OC = BC - OB = 25 - 9 = 16cm.

When the ABC is revolved about the hypotenuse, we get a double cone as shown in the attachment.

Volume of double cone = volume of the cone CAA` + volume of the cone BAA`

$\tt{ = \frac{1}{3} \pi { (OA)}^{2} \times OC + \frac{1}{3} \pi {(OA)}^{2} \times OB}$

$\tt{= \frac{1}{3} \pi \times {12}^{2} \times 16 + \frac{1}{3} \pi \times {12}^{2} \times 9}$

$\tt{= \frac{1}{3} \pi \times 144(16 + 9)}$

$\tt{= \frac{1}{3} \times 3.14 \times 144 \times 25 {cm}^{3}}$

$\tt{ = 3768 \: {cm}^{3} }$

Surface area of the doubled cone = (C.S.A. of cone CAA`) + (C.S.A. of cone BAA`)

$\tt{=\:(\pi\:×\:OA\:×\:AC)\:+\:(\pi\:×\:OA\:×\:AB)}$

$\tt{=\:(\pi\:×\:12\:×\:20)\:+\:(\pi\:×\:12\:×\:15)\:cm^2}$

$\tt{=\:420\:\pi\:cm^2}$

$\tt{=\:420\:×\:3.14cm^2}$

$\tt{=\:1320cm^2.}$

${\red{\boxed{I\:hope\:it\:will\:help\:you}}}$

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Answer 2

Step-by-step explanation:

When a right-angled triangle is revolved around its hypotenuse, a double cone is formed with same radius but with different heights.

it is given that, AB = 15 cm, AC = 20 cm

Let, OB = x and OA = y

Observe from the figure,

In right-angled triangle ABC,

By Pythagoras theorem [Hypotenuse²  = Base²  + Perpendicular² ]

BC²  = AC² + AB²

⇒ BC² = 20²  + 15²

⇒ BC²  = 400 + 225

⇒ BC² = 625

⇒ BC = 25 cm

In ΔOAB

AB² = OA²  + OB²

⇒ 15²  = x²  + y²  ……(1)

In ΔAOC

AC²  = OA²  + OC²

⇒ 20²  = y² + (BC – OB)²

⇒ 400 = y²  + (25 – x)²

⇒ 400 = y²  + 625 – 50x + x²

⇒ 400 = 15² + 625 – 50x

⇒ 400 = 225 + 625 – 50x

⇒ 50x = 450

⇒ x = 9 cm

from equation (1),

15²  = 9²  + y²

⇒ y² = 225 – 81

⇒ y² = 144

⇒ y = 12 cm

Also, OC = 25 – x = 25 – 13 = 12 cm²

Now, Volume of cone , V = 1/3 πr²h

Hence, volume of double cone 

= 1/3 π(OA)² × BO + 1/3 π(OA)² × OC

= 1/3 π(12)² × (OB + OC)

= 1/3 × 3.14 × 144 × 25

= 3768 cm³

Curved surface area of cone = πrl

Surface area of double cone = CSA of left cone + CSA of right cone

= π(OA)(AB) + π(OA)(AC)

= 3.14 × 12 × (15 + 20)

= 1318.8 cm²