A right angled triangle with sides 3 cm and 4 cm is revolved around its hypotenuse. Find the volume of the double cone thus generated.
Answers
Answer:
The Required volume of double cone is 30.17 cm³ .
Step-by-step explanation:
SOLUTION :
Here, ABC is a right angled triangle, right angled at A and BC is the hypotenuse.
Given :
AB = 3 cm
AC = 4 cm
In ∆ABC,
BC² = AB² + AC ²
[By Pythagoras theorem]
BC² = 3² + 4² = 9 + 16 = 25
BC² = 25
BC = √25
BC = 5 cm
As ∆ABC revolves about the hypotenuse BC. It forms two cones ABD & ACD.
In ∆AEB. & ∆CAB
∠AEB = ∠CAB = 90° &
∠ABE = ∠ABC = common
∆AEB ~ ∆CAB
[By, rule AA similarity criterion]
AE/CA = AB/BC
[In similar triangle sides are proportional]
AE /4 = ⅗
AE = 12/5 = 2.4 cm
Radius of the base of each cone, r = AE = 2.4 m
In right angled ∆AEB,
BE = √AB² - AE²
= √3² - 2.4² = √9 - 5.76 = √3.24 = 1.8 cm
BE = 1.8 cm
Height of the cone ABD = 1.8 cm
Height of the cone ACD = CE = BC - BE = 5 - 1.8 cm = 3.2 cm
Volume of the cone ABD,V1 = 1/3πr²h
V1 = ⅓ × 22/7 × 2.4² × 1.8
V1 = (22×10.368)/21 = 10.86 cm³
V1 = 10.86 cm³
Volume of the cone ABD = 10.86 cm³
Volume of the cone ACD,V2 = 1/3πr²h
V2 = ⅓ × 22/7 × 2.4² × 3.2
[Radius will be same as AD is common]
V2 = 405.504/21 = 19.31 cm³
V2 = 19.31 cm³
Volume of the cone ACD = 19.31 cm³
Required volume of double cone,V = V1 + V2
V = 10.86 cm³ + 19.31 cm³
V = 30.17 cm³
Hence, the Required volume of double cone is 30.17 cm³ .
HOPE THIS ANSWER WILL HELP YOU….
Answer:
The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.
Hypotenuse, AC =
3
2
+4
2
=
9+16
=
25
=5 cm
Area of ΔABC=
2
1
×AB×AC
⇒
2
1
×AC×DB=
2
1
×4×3
⇒
2
1
×5×DB=6
So, DB =
5
12
=2.4 cm
The volume of double cone = Volume of cone 1 + Volume of cone 2
=
3
1
πr
2
h
1
+
3
1
πr
2
h
2
=
3
1
πr
2
[h
1
+h
2
]=
3
1
πr
2
[DA+DC]
=
3
1
×3.14×2.4
2
×5
=30.14 cm
3
The surface area of double cone = Surface area of cone 1 + Surface area of cone 2
=πrl
1
+πrl
2
=πr[4+3]=3.14×2.4×7
=52.75 cm
2