Question

A right triangle whose sides are 3cm and 4cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.​

Answer 1

Given that in a right angled triangle, whose sides are 3 cm and 4 cm(other than hypotenuse).

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse .

Hypotenuse AC = Ö( 32 + 42) = 5 cm

Area of ΔABC = (1/2) x AB xAC  

⇒ (1/2) x AC x OB = (1/2)x 4 x3  

⇒ (1/2) x 5 x OB = 6  

⇒ OB = 12 / 5 = 2.4 cm.

Volume of double cone = Volume of cone 1 + Volume of cone 2

= (1/3) πr2h1 + (1/3)πr2h2

= (1/3) πr2(h1 + h2)  

= (1/3) πr2(OA +OC)  

= (1/3) x 3.14x(2.4)2 x(5)

= 30.14 cm3 surface area of double cone

= surface area of cone ABD + surface area of cone BCD

= πrl1 + πrl2  [where l1 and l2 are the slant heights of the cone ABD and BCD respectively]

thus the volume of the double cone is 30.14 cubic cm

and surface area of double cone is 52.8 sq cm

Answer 2

Answer :—

Volume of double Cone =

$\huge \: 30.17 \: {cm}^{2}$

&

Surface area of double Cone =

$\huge \: 52.8 \: cm^{2}$

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step by step explaination :—

Let, ABC be a right angled triangle, right angled at A and BC is the hypotenuse.{as shown on my attachment}

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Also, let AB = 3cm

Also, let AB = 3cmAlso, let AB = 3cm

and, AC = 4cm

Then,

$= > BC = \sqrt{3 {}^{2} + {4}^{2} } \\ </p><p> = > \: \: \: BC = \sqrt{9 + 16} \\ </p><p> = > \: \: \: \: \: \: \: \: \: \: BC = \sqrt{25} \\ </p><p> = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: BC = 5$

here, we get →

BC = 5cm

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As, ∆ABC revolves about the hypotenuse BC. It forms two cones ABD and ACD.{as shown in my attachment}

In ∆AEB and ∆CAB,

/_AEB = /_CAB ----{each 90°}

}/_ABE = /_ABC ----{common}

therefore, ∆AEB ~ ∆CAB --[by AA similarity]

$therefore - > \\ \frac{AE}{CA} = \frac{AB}{BC}--- [by, c.p.s.t]$

$=> \frac{AE}{4} = \frac{3}{4} \\ = > AE = \frac{3 \times 4}{5} \\ = > AE = \frac{12}{5} \\ => AE = 2.4$

So, radius of the base of each cone,

AE = 2.4cm

Now,

$= > \: BE = \sqrt{ {AB}^{2} - {AE}^{2} } \\ - by \: \: pythagoras \: \: theorem \\ = > \: \: \: BE = \sqrt{ {3}^{2} - {(2.4)}^{2} } \\ = > \: \: \: \: \: \: \: \: \: BE = \sqrt{9 - 5.76} \\ = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: BE = \sqrt{3.24} \\ \\ = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: BE = 1.8$

So, height of the cone ABD = BE = 1.8cm

therefore,

height of the cone, ACD = CE = BC — BE

CE = 5 — 1.8 = 3.2cm

Now,

Volume of the cone ABD →

$= \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{3} \pi {r}^{2} \\ = \frac{1}{3} \times \frac{22}{7} \times (2.4)^{2} \times 1.8 \\ = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{22}{21} \times 10.368 \\ \\ = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 10.86 {cm}^{2}$

similarly, volume of the cone ACD →

$= \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{3} \pi {r}^{2} \\ = \frac{1}{3} \times \frac{22}{7} \times (2.4)^{2} \times 3.2\\ (radius \: will \: be \: same \: as AD \\ is \: common) \: \: \\ = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{405.504}{21} \\ \\ = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 19.31{cm}^{2}$

therefore,

REQUIRED volume of double cone →

= 10.86 + 19.31

= 30.17cm³

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& also

surface area of cone ABD = πrl

$= \frac{22}{7} \times 2.4 \times 3= \frac{158.4}{7} \\ = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 22.63 {cm}^{2}$

similarly,

surface area of cone ACD = πrl

$= \frac{22}{7} \times 2.4 \times 4= \frac{211.2}{7} \\ = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 30.17 {cm}^{2}$

therefore,

required surface area of double cone →

= 22.63 + 30.17

= 52.8cm²

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hope it's helpful,,,,

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