Question

A semicircle sheet of paper of diameter 14cm is bent to form an open conical cup. Find the capacity of the cup

Answer 1

r=14/2=7cm

area of semi circle = CSA of cone

1/2πr^2=πrl

πr(1/2r)=πr(l)

1/2r=l

1/2×7=l

l=3.5

by pythagores theorem

h^2=p^2+b^2

p^2=h-^2-b^2

h^2=(3.5)^2-(7)^2

h=√-12.25+49

h=√36.75

h=6.06cm

volume =1/3πr^2h

1/3×22/7×7×7×6.06

=311.08cm³

Answer 2

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. $2\pi r =(\frac{1}{2})2\pi R$

$2r =R$

$2r =7$

$r =\frac{7}{2}$

Slant height of the conical cup = Radius of the semi-circular sheet

We know that,

$l^2= h^2+ r^2$

$(7)^2= h^2+ (\frac{7}{2})^2$

$49= h^2+\frac{49}{4}$

$h^2=49-\frac{49}{4}$

$h^2=\frac{196-49}{4}$

$h^2=\frac{147}{4}$

$h=\sqrt{\frac{147}{4}}$

$h=6.06$

Height of the conical cup = 6.06 cm

The capacity of the conical cup is given by,

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

$V= \frac{1}{3}\times \frac{22}{7}\times (\frac{7}{2})^2\times 6.06$

$V= \frac{1}{3}\times \frac{22}{7}\times \frac{49}{4}\times 6.06$

$V= 77.77\ cm^3$

Therefore, the capacity of the cup is 77.77 cm³.

#Learn more

A semicircular sheet of paper of diameter 28 cm is bent into an open conical cup. find the depth and the capacity of the cup

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