Question

using a thin metal sheet in the shape of a semicircle of sheet radius 48 cm a conical vessel is made find out the measurement of slant hieght and base radius of the vessel, calculate the curved surface area of the vessel, calculate hieght of the vessel, what is the ratio between the base radius and hieght and slant hieght,calculate the volume

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• Using a thin metal sheet in the shape of a semicircle of sheet radius 48 cm, a conical vessel is made. Find out the measurement of slant height and base radius of the vessel, calculate the curved surface area of the vessel, calculate height of the vessel. What is the ratio between the base radius and hieght and slant hieght? Calculate the volume.

Answer :-

Given:-

• Radius of semicircular sheet, r = 48 cm.

To find :-

• Slant height (l)
• Base Radius (R)
• Height of vessel (h)
• Curved Surface Area
• R : h : l
• Volume of vessel

Solution :-

• As we know that the circumference of a circle is given as-

• Circumference = πr

• Whereas, r is the radius of circle

☆ Now, Radius of circular sheet, r  =48 cm

∴ Circumference of semi - circular sheet = 48 π cm.

☆ When a semi-circular sheet is bent to form an open conical cup, the radius of the sheet becomes the slant height of the cup and the circumference of the sheet becomes the circumference of the base of the cone.

☆ Slant height of cup (l)= Radius of circular sheet = 48 cm

⇒ Slant height (l) of cone = 48 cm ----(1)

Now, Circumference of the base of cone = circumference of circular sheet = 48 π

Let R be the radius of the base of cone

∴2 π R = 48 π

⇒ R = 24 cm -------------(2)

☆ Let h be the height of cup.

Therefore, by using Pythagoras Theorem,

$\tt\implies \: {l}^{2} = {R}^{2} + {h}^{2}$

$\tt\implies \: {48}^{2} = {24}^{2} - {h}^{2}$

$\tt\implies \: {48}^{2} - {24}^{2} = {h}^{2}$

$\tt\implies \: {h}^{2} = (48 + 24)(48 - 24)$

$\tt\implies \: {h}^{2} = 72 \times 24$

$\tt\implies \: {h}^{2} = 24 \times 3 \times 24$

$\tt\implies \:h \: = 24 \sqrt{3} \: cm - - - (3)$

☆ Now, Curved Surface Area of cone = π R l

$\tt\implies \:Curved \: Surface \: Area = \dfrac{22}{7} \times 48 \times 24$

$\tt\implies \:Curved \: Surface \: Area = \dfrac{25344}{7} \: {cm}^{2}$

$\tt \: Now, \: R : h : l \\ \tt \:  \: 24 : 24 \sqrt{3} : 48 \\ \tt\implies \:1 : \sqrt{3} : 2$

$\tt \: Now, \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {R}^{2} h$

$\tt\implies \:Volume_{(cone)} = \dfrac{1}{3} \times \dfrac{22}{7} \times 24 \times 24 \times 24 \sqrt{3}$

$\tt\implies \:Volume_{(cone)} = \dfrac{101376 \sqrt{3} }{7} \: {cm}^{3}$

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More info:

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length ²+breadth ²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²