MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Volume of Combination of solids

Question 7The...

Question

Question 7
The curved surface area of a frustum of a cone is $π (r_{1} + r_{2}) l$ , where $l = \sqrt{h^{2} + (r_{1} + r_{2})^{2}}$ $r_{1} a n d r_{2}$ are the radii of the two ends of the frustum and h is the vertical height.

Open in App

Solution

The statement is false.
We know that if $r_{1} a n d r_{2}$ are the radii of the two ends of the frustum and h is the vertical height. Then curved surface area of a frustum is $π (r_{1} + r_{2}) l$ . where $l = \sqrt{h^{2} + (r_{1} - r_{2})^{2}}$ , where $l$ is the slant height of the cone.

Suggest Corrections

thumbs-up

1

Similar questions

Q. State True(1) or False(0).

The curved surface area of a frustum of a cone is

π l (r_{1} + r_{2})

l = \sqrt{h^{2} + {(r_{1} - r_{2})}_{1}^{2}}

r_{1} a n d r_{2}

are the radii of the two ends of the frustum and

h

is the vertical height.

Q. STATEMENT - 1: The slant height of frustum of a cone is 4 cm. and perimeters of its circular ends are 18 cm. and 6 cm. then the curved surface area is 48 sq. cm.
STATEMENT - 2: Curved surface area of frustum

= π (r_{1} + r_{2})

r_{1}

r_{2}

are radii of the frustum and

l

is a slant height.

Q. The radii of the ends of a frustum of a cone of slant height

l

r_{1}

r_{2}

. The curved surface area of the frustum of the cone is

Q. Question 5
Write ‘True’ or ‘False’ and justify your answer in the following
The volume of the frustum of a cone is

\frac{1}{3} π h [r_{1}^{2} + r_{2}^{2} - r_{1} r_{2}]

, where h is vertical height of the frustum and

r_{1}, r_{2}

are the radii of the ends.

Q. The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of the frustum complete the following activity.(

π = \frac{22}{7}

circumference₁ = 2

π

r₁= 132

r₁=

\frac{132}{π} = 134

circumference₂ =2

π

r₂= 88

r₂=

\frac{88}{2 π} = 134

slant height of frustum, l =

\sqrt{h^{2} + {(r_{1} - r_{2})}^{2}} \sqrt{1223^{2} + 1234^{2}} = 1234 cm

curved surface area of the frustum =

π

(r₁+ r₂ ) l

=

π \times 123 \times 132 = 123 sq . cm .

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Volume of Combination of solids

Standard X Mathematics

Join BYJU'S Learning Program

CrossIcon