A semicircular sheet of metal of diameter 14 cm is bent
to form an conical container. Find the capacity of the container .
Answers
Answer:
The capacity of the cup is 77.77 cm³.
Step-by-step explanation:
Given : A semicircle sheet of paper of diameter 14 cm is bent to form an open conical cup.
To find : The capacity of the cup ?
Solution :
A semicircle sheet of paper of diameter 14 cm.
The radius is R=7 cm.
Let radius and height of the conical cup be 'r' and 'h' respectively.
Circumference of the base of the cone = Length of arc of the semi-circle
i.e. 2\pi r =(\frac{1}{2})2\pi R2πr=(
2
1
)2πR
2r =R2r=R
2r =72r=7
r =\frac{7}{2}r=
2
7
Slant height of the conical cup = Radius of the semi-circular sheet
We know that,
l^2= h^2+ r^2l
2
=h
2
+r
2
(7)^2= h^2+ (\frac{7}{2})^2(7)
2
=h
2
+(
2
7
)
2
49= h^2+\frac{49}{4}49=h
2
+
4
49
h^2=49-\frac{49}{4}h
2
=49−
4
49
h^2=\frac{196-49}{4}h
2
=
4
196−49
h^2=\frac{147}{4}h
2
=
4
147
h=\sqrt{\frac{147}{4}}h=
4
147
h=6.06h=6.06
Height of the conical cup = 6.06 cm
The capacity of the conical cup is given by,
V= \frac{1}{3}\pi r^2 hV=
3
1
πr
2
h
V= \frac{1}{3}\times \frac{22}{7}\times (\frac{7}{2})^2\times 6.06V=
3
1
×
7
22
×(
2
7
)
2
×6.06
V= \frac{1}{3}\times \frac{22}{7}\times \frac{49}{4}\times 6.06V=
3
1
×
7
22
×
4
49
×6.06
V= 77.77\ cm^3V=77.77 cm
3
Therefore, the capacity of the cup is 77.77 cm³.
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The capacity of the container is 155.2 cubic cm.
Given: Semicircular metal sheet of diameter 14cm and convert into cone.
To Find : Volume of the conical container
Solution:
Radius of semicircle=7
The circumference of the semicircle is the circumference of the the cone
Circumference of the semicircle = 22/7 *7 = 22cm
Radius of cone =22=2*22/7r
r =7/2cm
Length of the cone = Radius of semicircle
Length of cone = 7cm
Height of cone =
49=49/4+ h2
h = 12.1 cm
Now volume of cone =
Volume of cone = 1/3 * 22/7 * 7/2 *7/2 *12.1
Volume of cone = 155.2 cubic cm.
Hence, the capacity of the cone is 155.2 cubic cm
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