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A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.)

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Short Answer Type

442.

A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm³ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?

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443.

In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

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444.

An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see the given Fig.

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Long Answer Type

445.
Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Let h be the height, l the slant height and r₁ and r₂ the radii of the circular bases of the frustum ABB’ A’ shown in Fig. such that r₁ > r₂.
Let the height of the cone VAB be h₁ and its slant height be i.e., VO = h₁ and VA = VB = l₁∴ VA’ = VA – AA’ = l₁– l
and VO’ = VO – OO’ = h₁– h
Here, ΔVOA ~ ΔVO‘A’

$therefore space space space fraction numerator VO over denominator VO apostrophe end fraction equals fraction numerator OA over denominator OA apostrophe end fraction equals fraction numerator VA over denominator VA apostrophe end fraction rightwards double arrow space space space fraction numerator straight h subscript 1 over denominator straight h subscript 1 minus straight h end fraction equals straight r over straight R equals fraction numerator straight l subscript 1 over denominator straight l subscript 1 minus 1 end fraction rightwards double arrow space space space space fraction numerator straight h subscript 1 minus straight h over denominator straight h subscript 1 end fraction equals straight F subscript 2 over straight F subscript 1 equals 1 minus 1 over straight l subscript 1 rightwards double arrow space space space straight h over straight h subscript 1 equals 1 minus straight r subscript 2 over straight r subscript 1 space and space straight l over straight l subscript 1 equals 1 minus straight r subscript 2 over straight r subscript 1 rightwards double arrow space space space space straight h over straight h subscript 1 equals fraction numerator straight r subscript 1 minus straight r subscript 2 over denominator straight r subscript 1 end fraction space and space straight l over straight l subscript 1 equals fraction numerator straight r subscript 1 minus straight r subscript 2 over denominator straight r subscript 1 end fraction rightwards double arrow space space space space space straight h subscript 1 equals fraction numerator hr subscript 1 over denominator straight r subscript 1 minus straight r subscript 2 end fraction space and space straight l subscript 1 equals fraction numerator lr subscript 1 over denominator straight r subscript 1 minus straight r subscript 2 end fraction space space space space... left parenthesis straight A right parenthesis$

Let h be the height, l the slant height and r1 and r2 the radii of

Now,
Height of the cone VA‘B’

$equals space straight h subscript 1 minus straight h equals fraction numerator hr subscript 1 over denominator straight r subscript 1 minus straight r subscript 2 end fraction minus straight h equals fraction numerator hr subscript 2 over denominator straight r subscript 1 minus straight r subscript 2 end fraction space space space space... left parenthesis straight B right parenthesis$
Slant height of the cone VA‘B’

$equals space straight l subscript 1 minus straight l space equals fraction numerator lr subscript 1 over denominator straight r subscript 1 minus straight r subscript 2 end fraction minus straight l equals fraction numerator lr subscript 2 over denominator straight r subscript 1 minus straight r subscript 2 end fraction space space space space space space.... left parenthesis straight C right parenthesis$

Let S denote the curved surface area of the frustum of cone. Then,
S = Lateral (curved) surface area of cone VAB
- Curved surface area of cone VA‘B’

$rightwards double arrow space space space space space space space space straight S space equals space πr subscript 1 straight l subscript 1 minus πr subscript 2 left parenthesis straight l subscript 1 minus straight l right parenthesis rightwards double arrow space space space space space space space space straight S space equals space πr subscript 1. fraction numerator lr subscript 1 over denominator straight r subscript 1 minus straight r subscript 2 end fraction minus πr subscript 1. fraction numerator lr subscript 2 over denominator straight r subscript 1 minus straight r subscript 2 end fraction$
[Using (A) and (C)]

$rightwards double arrow space space space space straight S space equals space straight pi space open parentheses fraction numerator straight r subscript 1 squared minus straight r subscript 2 squared over denominator straight r subscript 1 minus straight r subscript 2 end fraction close parentheses straight l rightwards double arrow space space space space space space space equals space straight pi space left parenthesis straight r subscript 1 plus straight r subscript 2 right parenthesis space straight l$

Curved surface area of the frustum
= π(r₁ + r₂)l
Total surface area of the frustum
= Lateral (curved) surface area
+ Surface area of circular bases
= π (r₁ + r₂) I + πr₁² + πr₂²= π {(r₁ + r₂) l + r₁² + r₂²}.

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Short Answer Type

446.

Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

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447. The radius and height of a cylinder are in the ratio 5:7 and its volume is 550 cm³. Find its radius.

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448. The curved surface area of height circular cone is 12320 cm². If the radius of the base is 56 cm, find its height.

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449. A conical flask is full of mater. The flask was base m radius r and height n. the mater is poored into a cylindrical flask of base radius mr. find the height of water in the cylindrical flask.

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