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 a double cone formed.

Answer 1

let ABC be a right angle triangle.

• In triangle ABC, angle A = 90°

• And BC is the hypotenuse.

[see the attached image]

Now,

AB = 3cm

And AC = 4cm

We can get BC by Pythagoras theorem.

⇒ BC² = AB² + AC²

⇒ BC² = 3² + 4²

⇒ BC² = 9 + 16

⇒ BC² = 25

⇒ BC = √25

⇒BC = 5cm

Now,

According to the question:

The ΔABC revolves around the hypotenuse BC. It forms two cones ABD and ACD.

SO,

By AA(Angel-Angle) similarity criterion,

ΔAEB and ΔCAB are similar.

[∵ ∠AEB = CAB = 90° and ∠ABE = ∠ABC]

$\boxed{\frac{AE}{CA}=\frac{AB}{BC}}$  

[Corresponding sides are proportional in similar triangles]

$\implies \frac{AE}{4}=\frac{3}{5}$

$\implies AE = \frac{12}{5} = \boxed{2.4cm}$

So,

The radius of the base of each cone,

AE = 2.4cm

Now,

In right-angled ΔAEB,

⇒ BE² = AB² - AE²  [Pythagoras Theorem]

⇒ BE² = 3² - (2.4)²

⇒ BE² = 9 - 5.76

⇒ BE² = 3.24

⇒ BE = √3.24

⇒ BE = 1.8 cm

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

Also,

∴ Height of the cone ACD = CE = BC - BE

= 5 - 1.8 = 3.2cm

Now,

the volume of the cone ABD,

$= \frac{1}{3} \pi rh$

$= \frac{1}{3} \times \frac{22}{7} \times (2.4)^{2} \times (1.8)$

$=\frac{22}{21} \times 10.368 = \boxed{\sf 10.86cm^{2}}$

And,

The volume of the cone ACD,

$= \frac{1}{3} \pi r^{2}h$

$=\frac{1}{3} \times \frac{22}{7} \times (2.4)^{2} \times 3.2$

[ The radius will be the same as AD is common]

$= \frac{405.504}{21} = \boxed{\sf 19.31cm^{3}}$

So,

$\underline{\red{\rm Required\ volume\ of\ double\ cone - }}$

= 10.86 + 19.31

= 30.17 cm cube.

$Now,$

The surface area of the cone ABD

= πrl unit sq.

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

And the surface area of the cone ACD

= πrl unit sq.

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

Hence,

The required surface area of the double cone

= 22.63 + 10.17

= 52.8cm sq.

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}

__________________

Also, let AB = 3cm

Also, 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 →

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°}

/_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)cm = 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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