WBBSE Solutions For Class 10 Maths Mensuration Chapter 4 Right Circular Cone

WBBSE Solutions For Class 10 Maths Mensuration Chapter 4 Right Circular Cone

What is a right circular cone?

Right circular cone

If a right-angled triangle is rotated by taking one of its sides as the axis except the hyponetuse with ah angle of 360°, then the solid object thus produced is known as right circular cone.

For example, a comet of Muri, a cone of ice cream, etc.

The right-angular point of the right-angled triangle is O and ∠AOB = 90°.

Now, keeping the right-angular point O fixed and taking AO as the axis if the triangle is rotated with an angle of 360°, then the right circular angle ABC is produced.

The line segment joining the vertex and the centre of the circular base will be perpendicular to the base of the cone.

The line segment joining the vertex and the centre of the circular base is not perpendicular to the base of the cone.

WBBSE Solutions for Class 10 Maths

So it is not a right circular cone.

Remember that a closed right circular cone has two surfaces-one is the curved surface and the other is the plane surface.

Some essential definition regarding a right circular cone:

Axis: The side, of a right-angled triangle adjacent to the right angle around which the triangle is rotated with an angle of 360°, a right circular cone is produced is biown as the axis of the cone.

The axis of the right circular cone is AO.

Base: The circular plane, which is produced by the other side of the right-angled triangle adjacent to the right angle, when the triangle is rotated with an angle of 360°, is known as the base of the right-circular cone.

The base of a right circular cone is a circle. O is the centre of the circular base.

Vertex: The point, at the top of the right circular cone which remains fixed while producing the cone by rotating the right-angled triangle with jan angle of 360° except the right-angular point is known as the vertex of the right circular cone.

A is the vertex of the right circular cone.

The radius of the base: The base of the right circular cone is-a circle and the radius of this circle is known as the radius of the base.

OB is the radius of the base of the right-circular cone.

Height: The perpendicular distance between the centre of the base of the right circular cone and the vertex of the cone is known as the height of the right circular cone.

AO is the height of the right circular cone AOB.

Slant height: The length of the hypotenuse of the right-angled triangle which by rotation produces the cone is known as the slant height of the right circular cone.

AB is the slant height of the right circular cone.

Semi-vertical angle: The angle between the axis of a right circular cone and its slant height is known as the semi-vertical angle.

∠OAB is the semi-vertical angle.

Relation among the slant height, height and the radius of the base of a right circular cone:

Let ABC be the right circular cone produced by rotating the right-angled triangle AOB by keeping the side AO fixed and with an angle of 360°.

Here, the height of the cone = AO and the radius of it = OB. The slant height of the cone = AB.

∴ by Pythagoras theorem, AB² = OA² + OB² i.e.,

(slant height)² = (height)² + (radius of the base)²

Hence if the slant height, height and the radius of the base be l, h and r Unit respectively, then l² = h² + r²

or, l = \(\sqrt{h^2+r^2}\) and r = \(\sqrt{h^2+r^2}\) and h = \(\sqrt{l^2-r^2}\)

Formulas related to right circular cone:

Let the radius of the base of a right circular cone is r unit, height h unit and slant height l unit. Then

1. The lateral surface area of the right circular cone

= \(\frac{1}{2}\) x circumference of the circular base x slant height = \(\frac{1}{2}\) x 2πr x l sq-unit = πrl sq-unit

2. The area of the base of the right circular cone = area of the circular base = πr² sq-unit.

3. The total surface area of the right circular cone = lateral surface area + area of the base = (πrl + πr²) sq-unit = πr (l + r) sq-unit.

4. The volume of the right circular cone = \(\frac{1}{3}\)πr²h cubic-unit.

Mensuration Chapter 4 Right Circular Cone Multiple Choice Questions

“WBBSE Class 10 right circular cone solved examples”

Example 1. The ratio of the volumes of two right circular cones is 1: 4 and the ratio of their lengths of radii of the bases is 4: 5. Then the ratio of their heights is

1. 1: 5
2. 5: 4
3. 25: 16
4. 25: 64

Solution:

Given

The ratio of the volumes of two right circular cones is 1: 4 and the ratio of their lengths of radii of the bases is 4: 5.

Let the lengths of the radii of the bases of the right circular cone are 4x unit and 5x unit.

Also, let the heights of the two right circular cones are h₁ unit and h₂ unit.

So, the volume of the cones are \(\frac{1}{3}\)π(4x)² h₁ cubic-units and \(\frac{1}{3}\)π(5x)² h₂ cubic-unit.

As per question, \(\frac{1}{3}\)π(4x)² h₁ : \(\frac{1}{3}\)π(5x)² h₂ cubic-unit.

⇒ 16 x² h₁ : 25 x² h₂ = 1: 4

⇒ \(\frac{16 x^2 h_1}{25 x^2 h_2}=\frac{1}{4}\)

⇒ \(\frac{h_1}{h_2}\) = \(\frac{1}{4}\times \frac{25}{16}\)

⇒ \(\frac{h_1}{h_2}=\frac{25}{64}\)

⇒ h₁: h₂ = 25: 64

Hence the ratio of the heights of the two right circular cones is 25: 64.

∴ 4. 25: 64 is correct.

The ratio of their heights is  25: 64

Example 2. If keeping the length of radius of a right circular cone fixed, the height of it is increased by 2 times, then the volume of cone will be increased by

1. 100 %
2. 200 %
3. 300 %
4. 400 %

Solution:

Given

If keeping the length of radius of a right circular cone fixed, the height of it is increased by 2 times

Let the radius of the right circular cone be r unit and height is h unit.

∴ if the volume of the cone be V cubic-unit, then V = \(\frac{1}{3}\)πr²h.

Now, if the height of the cone be doubled keeping, the radius of it fixed, then the new volume of the cone

V₁ = \(\frac{1}{3}\)πr² (2h) = 2. \(\frac{1}{3}\)πr²h = 2v…….(1)

So, the volume increases = (V₁ – V) cubic-unit = (2V – V) cubic-unit = V cubic-unit.

Hence the percent of the increment in volume

= \(\frac{V}{V}\) x 100%\(\left[\frac{\text { increment in volume }}{\text { primary volume }} \times 100 \%\right]\) = 100%

∴ 1. 100% is correct

The volume of cone will be increased by 100%

Example 3. If each of the radius and height of a right circular cone be doubled, then the volume of the cone will be

1. 3 times
2. 4 times
3. 6 times
4. 8 times

Solution:

Given

If each of the radius and height of a right circular cone be doubled,

If the radius, height and volume of the cone be r unit, h unit and V cubic-unit respectively, then V = \(\frac{1}{3}\)πr²h .

Now, if the volume of the cone be V₁ cubic-unit when its radius and height be doubled, then

V₁ = \(\frac{1}{3}\)π(2r)².2h = \(\frac{1}{3}\)π.4r² .2h = 8.\(\frac{1}{3}\)πr² h = 8V

∴ If the radius arid height of a right circular cone be doubled its volume will be 8 times of the previous volume.

∴ 4. 8 times is correct.

The volume of the cone will be 8

Example 4. If the length of radius of a right circular cone be \(\frac{r}{2}\) unit and its slant height be 2l unit, the total surface area of it will be

1. 2πr (l + r) sq-unit
2. πr (l + r) sq-unit
3. \(\pi r\left(1+\frac{r}{4}\right)\) sq-unit
4. 2πrl sq-unit

Solution:

Given

The radius of the right circular cone is \(\frac{r}{2}\), slant height = 2l unit

∴ the total surface area of the cone = \(\pi \times \frac{r}{2}\left(\frac{r}{2}+2 l\right)\)sq- unit

= \(\pi \times \frac{r}{2} \times 2\left(\frac{r}{4}+l\right)\)sq- unit

= \(\pi r\left(l+\frac{r}{4}\right)\) sq-unit

∴ 2. πr (l + r) sq-unit is correct.

The total surface area of it will be πr (l + r) sq-unit

“Mensuration problems on right circular cone for Class 10”

Example 5. If a triangle of sides 6 cm, 8 cm and 10 cm be rotated keeping the side of length 8 cm fixed, then the volume of the cone thus produced will be

1. 96 π cc
2. 120 π cc
3. 128 π cc
4. 200 π cc

Solution:

Given

If a triangle of sides 6 cm, 8 cm and 10 cm be rotated keeping the side of length 8 cm fixed,

Since 6² + 8² = 10²,

∴ the given triangle is a right-angled triangle, the hypotenuse of which is 10 cm.

Now if the triangle is rotated keeping the side of length 8 cm fixed, then height of the cone will be 8 cm and the radius of its base will be 6 cm

[∵ the other side of length 10 cm is the hypotenuse]

∴ the volume of the cone = \(\frac{1}{3}\)π x 6² x 8 cc = 96π cc

∴ 1. 96 π cc is correct.

The volume of the cone thus produced will be 96 π cc

Example 6. The area of the base of a right-circular cone is 25 sq-cm and the height is 10 cm. Then the volume of the cone will be

1. \(\frac{150}{3}\) cc
2. \(\frac{250}{3}\) cc
3. \(\frac{350}{3}\) cc
4. \(\frac{400}{3}\) cc

Solution:

Given

The area of the base of a right-circular cone is 25 sq-cm and the height is 10 cm.

The volume of the cone = \(\frac{1}{3}\) x area of the base x height = \(\frac{1}{3}\) x 25 x 10 cc = \(\frac{250}{3}\) cc

∴ 2. \(\frac{250}{3}\) cc is correct.

The volume of the cone will be \(\frac{250}{3}\) cc

Example 7. If the numerical values of the area of the base and the volume of a right circular cone of radius 4 cm be equal, then the slant height of the cone is

1. 3 cm
2. 4 cm
3. 5 cm
4. 6 cm

Solution:

Given

If the numerical values of the area of the base and the volume of a right circular cone of radius 4 cm be equal,

Let the height of the cone be h cm.

As per question, \(\frac{1}{3}\)π .4² .h = π . 4² ⇒ h = 3

∴ the slant height of the cone = \(\sqrt{4^2+3^2}\)

∴ 3. 5 cm is correct.

The slant height of the cone is 3. 5 cm

“Chapter 4 right circular cone exercises WBBSE solutions”

Example 8. If the height of a right circular cone be increased by 10% keeping its radius fixed, then the volume of the cone will be

1. 5%
2. 10%
3. 15%
4. 20%

Solution:

Given

If the height of a right circular cone be increased by 10% keeping its radius fixed

Let the radius of the base of a right circular cone be r unit and its height be h unit.

∴ if the volume of the cone be V cubic-unit, then V = \(\frac{1}{3}\)πr²h…….(1)

Now, if the height be increased by 10% then the height will be

\(\left(h+h \times \frac{10}{100}\right)\) unit = \(\left(h+\frac{h}{10}\right)\) unit = \(\frac{11 h}{10}\) unit

Then the volume of the cone = \(\frac{1}{3}\)πr² x \(\frac{11 h}{10}\) cubic-unit = \(\frac{11 h}{10}\) x \(\frac{1}{3}\)πr²h cubic -unit

= \(\frac{11 V}{10}\) cubic -unit [by(1)]

∴ the icrement of volume = \(\left(\frac{11 \mathrm{~V}}{10}-\mathrm{V}\right)\) cubic- unit = \(\frac{V}{10}\) cubic- unit

So, the percent of increment of volume = \(\frac{\frac{\mathrm{V}}{10}}{\mathrm{~V}}\) x 100% = 10%

∴ 2. 10% is correct.

The volume of the cone will be 10%

Example 9. The volume of a right circular conical tent is 1232 cc. If the height of the tent be 24 metres, the area of the base of the cone will be

1. 140 sq-m
2. 145 sq-m
3. 154 sq-m
4. 160 sq-m

Solution:

Given

The volume of a right circular conical tent is 1232 cc. If the height of the tent be 24 metres,

Let the radius ofthe base of the tent be r metre.

∴ the volume of the tent = \(\frac{1}{3}\)πr² x 24 cc

∴ \(\frac{1}{3} \times \frac{22}{7} \times r^2\) x 24 = 1232

or, \(r^2=\frac{1232 \times 3 \times 7}{22 \times 24}\)

∴ the area of the base of tent = nr2 sq-cm = \(\frac{22}{7} \times \frac{1232 \times 7 \times 3}{22 \times 24}\) sq-cm = 154 sq-cm.

∴ 3. 154 sq-m is correct.

The area of the base of the cone will be 154 sq-m

Example 10. The difference of the square of the height and the square of the base of a right circular cone of slant height 10 cm is 28 cm, then the volume of the cone will be

1. 42 π cc
2. 54 π cc
3. 72 π cc
4. 96 π cc

Solution:

Given

The difference of the square of the height and the square of the base of a right circular cone of slant height 10 cm is 28 cm,

Let the height of the right circular cone is h cm and the radius of the base be r cm.

The slant height of the cone is 10 cm.

∴ h² + r² = (10)² or, h² + r² = 100…..(1)

Again, h² – r² = 28…….(2)

Adding (1) and (2) we get, 2h² = 128 or, h² = 64 or, h = 8.

r² = 100 – h² = 100 – 64 = 36.

∴ the volume of the cone = \(\frac{1}{3}\)π x n x 36 x 8 cc = 96πcc.

∴ 4. 96 cc is correct.

The volume of the cone will be 96 cc.

Mensuration Chapter 4 Right Circular Cone True Or False

“Class 10 Maths volume of right circular cone problems”

Example 1. If the radius of the base of a right circular cone be halved and its height be doubled, then the volume of the cone remains the same.

Solution:

Given:

If the radius of the base of a right circular cone be halved and its height be doubled

If the volume of a right circular cone of radius r unit and height h unit be V cubicunit, then V = \(\frac{1}{3}\)πr²h

Now, if the radius of its base be halved and its height be doubled, then the volume of the cone

= \(\frac{1}{3} \pi\left(\frac{r}{2}\right)^2\).2h cubic -unit

= \(\frac{1}{3} \pi \cdot \frac{r^2}{4}\).2h cubic- unit

= \(\frac{1}{2}\).\(\frac{1}{3}\)πr²h cubic- unit = \(\frac{1}{2}\) V

i.e., the volume is halved.

Hence the given statement is false.

Example 2. The slant height of a right circular cone is always greater than the height of the cone.

Solution:

The height, radius and slant height of right circular cone are the three sides of a rightangled triangle.

Since slant height is the hypotenuse of this right-angled triangle, so it is always greater than its height.

Hence the given statement is true.

Example 3. Among the total surface area, lateral surface area and the area of the base of a right circular cone, the total surface area is the greatest in magnitude.

Solution:

Given:

Among the total surface area, lateral surface area and the area of the base of a right circular cone

Since for a right circular cone,

the total surface area = (area of lateral surface) + (area of base),

so, the total surface area is the greatest.

Hence the given statement is true.

Example 4. If the area of the base of a right circular cone be 3 times of its volume, then the height of it will be 1 unit.

Solution:

Given:

If the area of the base of a right circular cone be 3 times of its volume

Let the radius of the right circular cone be r unit and its height is h unit.

As per question, πr² =3 x \(\frac{1}{3}\)πr²h ⇒ h = 1

∴ the height of the cone = 1 unit.

Hence the given statement is true.

Example 5. If the area of the base of a right circular cone be x sq-unit and the volume be y cubic-unit, then the height of the cone will be \(\frac{3x}{y}\) unit.

Solution:

Given:

If the area of the base of a right circular cone be x sq-unit and the volume be y cubic-unit

We know, volume = \(\frac{1}{3}\) x area of base x height

=» height = \(\frac{\text { volume } \times 3}{\text { area of base }}\)

= \(\frac{y \times 3}{x}\)unit = \(\frac{3 y}{x}\)unit.

Hence if the area of the base be x sq-unit and its volume be y cubic-unit, the height will be \(\frac{3 y}{x}\) unit.

Hence the given statement is true.

Mensuration Chapter 4 Right Circular Cone Fill In The Blanks

Example 1. AC is the hypotenuse of a right-angled triangle ABC. The radius of the right circular cone which is produced by one complete rotation of the triangle with the side AB as the axis is _______

Solution:

Given:

AC is the hypotenuse of a right-angled triangle ABC. The radius of the right circular cone which is produced by one complete rotation of the triangle with the side AB as the axis i

BC

Since the hypotenuse of the right-angled triangle is AC and the right circular cone is produced by one complete rotation of the triangle keeping the side AB fixed, so AB is the height of the cone.

Clearly, the radius of the base is BC.

Example 2. If the volume of a right circular cone be V cubic-unit and the area ofthe base be A sq-unit, then the height of the cone is ______

Solution:

Given:

If the volume of a right circular cone be V cubic-unit and the area ofthe base be A sq-unit

\(\frac{3 \mathrm{~V}}{\mathrm{~A}}\)

Since volume = \(\frac{1}{3}\) x area of the base x height

∴ V = \(\frac{1}{3}\) x A x h

⇒ h = \(\frac{3 \mathrm{~V}}{\mathrm{~A}}\)

The height of the cone is = \(\frac{3 \mathrm{~V}}{\mathrm{~A}}\)

Example 3. The number of surfaces of a closed right pircular cone is ______

Solution: 2

Example 4. The number of vertices of a right circular cone is ______

Solution: 1

Example 5. The lateral surface area of a right circular cone = (Total surface area) – ________

Solution: area of base

Mensuration Chapter 4 Right Circular Cone Short Answer Type Questions

“Understanding right circular cone in Class 10 Maths”

Example 1. The height of a right circular cone is 12 cm and its volume is 100 n cc, then find the radius of the cone.

Solution:

Given:

The height of a right circular cone is 12 cm and its volume is 100 n cc

Let the radius of the right circular cone be r cm and its height is 12 cm.

∴ the volume of the cone = \(\frac{1}{3}\)πr² x 12cc = 4πr²cc

As per question, 4πr² = 100π or, r² = 25, or, r = 5

Hence the radius of the cone = 5 cm.

Example 2. The curved surface area of a right circular cone is V5 times of its base area. Find the ratio of the height and the length of radius of the cone.

Solution:

Given:

The curved surface area of a right circular cone is V5 times of its base area.

Let the radius of the cone is r unit, height is h unit and slant height is l unit.

∴ the curved surface area of the cone = πrl sq-unit = πr\(\sqrt{h^2+r^2}\) sq- unit.

Also, the area of the base = πr² sq-unit.

As per question, πr\(\sqrt{h^2+r^2}\) = √5πr² or, \(\sqrt{h^2+r^2}\) = √5r or, h² + r² = 5r² or, h² = 4r²

or, h² = (2r)² or, h = 2r or, \(\frac{h}{r}=\frac{2}{1}\) or, h : r = 2 : 1.

Hence the ratio of the height and radius of the right circular cone is 2: 1.

Example 3. If the volume of a right circular cone is V cubic-unit, base area is A sq-unit and height is H unit, then find the value of \(\frac{AH}{V}\)

Solution:

Given:

If the volume of a right circular cone is V cubic-unit, base area is A sq-unit and height is H unit

Let the radius of the right circular cone is r unit. the volume of the cone,

V = \(\frac{1}{3}\)πr²H cubic- unit,

Again, the area of the base of the cone, A = πr² sq-unit

∴ \(\frac{\mathrm{AH}}{\mathrm{V}}=\frac{\pi r^2 \mathrm{H}}{\frac{1}{3} \pi{ }^2 \mathrm{H}}\)

⇒ \(\frac{\mathrm{AH}}{\mathrm{V}}=3\)

Hence the value of \(\frac{AH}{V}\) = 3

Example 4. The numerical values of the volume and the lateral surface area of a right circular cone are equal. If the height and the radius of the cone are h unit and r unit respectively, then find the value of \(\frac{1}{h^2}+\frac{1}{r^2}\).

Solution:

Given:

The numerical values of the volume and the lateral surface area of a right circular cone are equal. If the height and the radius of the cone are h unit and r unit respectively

Let the slant height of the right circular cone is l unit.

Also, the height and radius of the cone are h and r unit respectively.

As per question, \(\frac{1}{3}\)πr²h = πrl

or, rh = 3l

⇒ r²h²  = 9l²  

⇒ r² h² = 9 (h² + r²)

⇒ \(\frac{h^2+r^2}{r^2 h^2}=\frac{1}{9}\)

⇒ \(\frac{h^2}{r^2 h^2}+\frac{r^2}{r^2 h^2}=\frac{1}{9}\)

⇒ \(\frac{1}{r^2}+\frac{1}{h^2}=\frac{1}{9}\)

⇒ \(\frac{1}{h^2}+\frac{1}{r^2}=\frac{1}{9}\)

Hence the value of \(\frac{1}{h^2}+\frac{1}{r^2}=\frac{1}{9}\)

“WBBSE Mensuration Chapter 4 practice questions on cones”

Example 5. The ratio of the lengths of the radii of the bases of a right circular cylinder and of a right circular cone is 3: 4 and the ratio of their heights is 2 : 3; Find the ratio of the volumes of the cylinder and the cone.

Solution:

Given:

The ratio of the lengths of the radii of the bases of a right circular cylinder and of a right circular cone is 3: 4 and the ratio of their heights is 2 : 3

Let the radii of the bases of the right circular cylinder and the right circular cone are 3r units and 4r units.

Also, let the heights of the cylinder and, the cone are 2h units and 3h units.

∴ the ratio of the volumes of the right circular cylinder and the right circular cone will be

\(\pi(3 r)^2 \cdot 2 h: \frac{1}{3} \pi(4 r)^2 \cdot 3 h=\pi \cdot 9 r^2 \cdot 2 h: \frac{1}{3} \pi \cdot 16 r^2 \cdot 3 h\)

= \(\frac{18 \pi r^2 h}{16 \pi r^2 h}\) = 18: 16 = 9: 8

Hence the required ratio is 9 : 8.

Example 6. The ratio of the radius of the base of a right circular cone and its height is 3: 7; The volume of the cone is 528 cc. Find the diameter of the cone.

Solution:

Given:

The ratio of the radius of the base of a right circular cone and its height is 3: 7; The volume of the cone is 528 cc.

Let the radius of the right circular cone and its height be 3x cm and 7x cm.

.-. the volume-of the cone = \(\frac{1}{3}\)π(3x)².7x cubic- unit

As per question, \(\frac{1}{3}\)π(3x)² . 7x = 528 or, \(\frac{1}{3}\)π x 9x² . 7x = 528

or, \(\frac{1}{3} \times \frac{22}{7} \times 63 x^3\) = 528

or, x³ = \(=\frac{528 \times 3 \times 7}{22 \times 63}\) or, x³ = 8 or, x = 2

∴ the radius of the cone = 3 x 2 cm = 6 cm

∴ the diameter of the cone = 2 x 6 cm = 12 cm

Hence the diameter of the cone =12 cm.

Example 7. The radius of a right circular cone is 7 cm at its vertical angle is 60°. Find the curved surface area of the cone.

Solution:

Given:

The radius of a right circular cone is 7 cm at its vertical angle is 60°.

The semi-vertical angle is \(\frac{60^{\circ}}{2}\) = 30°.

Now, from the right-angled ΔAOB, we get,

\(\frac{\mathrm{AB}}{\mathrm{OB}}\) = cosec 30° or, \(\frac{\mathrm{AB}}{7}\) = 2

[∵ OB = radius = 7 cm and cosec 30° = 2]

or, AB = 14

∴ the height of the cone = 14 cm.

∴ the curved surface area of the cone = \(\frac{22}{7}\) x 7 x 14sq-cm = 308 sq-cm.

Hence the curved surface area of the cone = 308 sq-cm.

Example 8. If the height and the area of base of a right circular cone be increased by 4 times, then how many times the volume of the cone will be increased?

Solution:

Given:

If the height and the area of base of a right circular cone be increased by 4 times

Let the initial radius of the base and height of the cone be r unit arid h unit respectively.

Now, if the area of the base be increased by 4 times, let the radius of the base will be r₁ unit.

∴ \(\pi r_1^2=4 \pi r^2\)

⇒ \(r_1^2=4 r^2\)

⇒ \(r_1^2=(2 r)^2\)

⇒ r₁ = 2r

If the initial volume of the cone be V cubic-unit, and the volume after the increment be V₁ cubic-unit, then

V = \(\frac{1}{3}\)πr²h and V₁ = \(\frac{1}{3}\)π .(2r)² .4h = 16.\(\frac{1}{3}\)πr²h = 16V. [∵ V = \(\frac{1}{3}\)πr²h]

Hence the volume of the cone will be 1 6 times of its initial volume, if its height and area of base is increased by 4 times.

Example 9. The curved surface area of a right circular cone is 42 times of the area of the base. Then what is the vertical angle of the cone?

Solution:

Given:

The curved surface area of a right circular cone is 42 times of the area of the base.

Let the radius of the base of the right circular cone be r unit and its slant height be l unit.

As per question, πrl = √2πr² ⇒ l = √2r

Now, let the semi-vertical angle be θ.

∴ from the right-angled ΔAOB we get, \(\frac{OB}{AB}\)= sinθ

⇒  sinθ = \(\frac{r}{\sqrt{2} r}=\frac{1}{\sqrt{2}}\) = sin45°

⇒ θ = 45°.

Hence the vertical angle = 2 x 45° = 90°.

Example 10. The area of bases of two right circular cones are equal. If the ratio of the slant heights be 2 : 3, then what will be the ratio of their curved surface areas?

Solution:

Given:

The ared of bases of two right circular cones are equal. If the ratio of the slant heights be 2 : 3

The area of the bases of the right circular cones are equal.

The radii of the bases of two cones will be also be equal.

Let the equal radii of the two cones be r unit and their slant heights be 21 units and 3l units respectively.

So, the ratio of their curved surface areas = πr2l: πr3l = 2:3.

Hence the required ratio of their curved surface areas is 2 : 3.

Mensuration Chapter 4 Right Circular Cone Long Answer Type Questions

“Step-by-step solutions for right circular cone Class 10”

Example 1. If the height and slant height of a cone are 6 cm and 10 cm respectively. Then determine the total surface area and the volume of the cone.

Solution:

Given:

If the height and slant height of a cone are 6 cm and 10 cm respectively

The height of the right circular cone is 6 cm and its slant height is 10 cm.

Let the radius of the cone be r cm.

∴ 6² + r² = 10² ⇒ r² = 100 – 36 = 64 ⇒ r = √64=8

∴ the radius of the cone is 8 cm.

∴ the total surface area of the cone = π x 8 (8 + 10) sq-cm = \(\frac{22}{7}\) x 8 x 18 sq-cm = 144π sq-cm.

Again, the volume of the cone = \(\frac{1}{3}\)π x (8)² x 6 cubic- cm = 128π cubic- cm.

Hence the total surface area of the cone = 144 n sq-cm and volume = 128 7t cubic-cm.

Example 2. 77 sq-cm tripal is required to make a right circular conical tent. If the slant height of the tent is 7 m, then calculate the base area of the tent.

Solution:

Given:

77 sq-cm tripal is required to make a right circular conical tent. If the slant height of the tent is 7 m,

The slant height of the tent is 7 metres.

Let the radius of the base of the tent be r metres.

∴ the curved surface area of the tent = \(\frac{22}{7}\) x r x 7 sq-m = 22r sq-m.

As per question, 22r = 77 ⇒ r = \(\frac{77}{22}=\frac{7}{2}\)

∴ the area of the base of the tent = \(\frac{22}{7} \times\left(\frac{7}{2}\right)^2\) sq-m = \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) sq-m

= \(\frac{77}{2}\) sq- m = 38.5 sq- m.

∴ the required base area of the tent = 38.5 sq-m.

Example 3. The length of the base diameter of a wooden toy of conical shape is 10 cm. The expenditure for polishing whole surfaces of the toy at the rate of 2.10 per square-metre is 429. Calculate the height of the toy. Also determine the quantity of wood which is required to make the toy.

Solution:

Given:

The length of the base diameter of a wooden toy of conical shape is 10 cm. The expenditure for polishing whole surfaces of the toy at the rate of 2.10 per square-metre is 429.

The base diameter of the toy = 10 cm.

∴ Radius of the base of the toy = \(\frac{10}{2}\) cm = 5cm.

Let the slant height of the toy = l cm

∴ the curved surface area of the toy = \(\frac{22}{7}\) x 5 x / sq-cm

₹2.10 is expent to polish 1 sq-m of the area.

∴ ₹1 is expent to polish \(\frac{1}{2 \cdot 10}\) sq-m of the area.

∴ ₹429 is expent to polish \(\frac{1 \times 429}{2 \cdot 10}\) sq-m of the area.

As per condition, \(\frac{22}{7}\) x 5 x l = \(\frac{429}{2 \cdot 10}\)

⇒ l = \(=\frac{429 \times 7}{2 \cdot 10 \times 22 \times 5}\) ⇒ l = 13

So, the slant height of the toy = 13 cm.

Let the height of the toy be h cm

∴ h² + 5² = (13)² or, h² + 25 = 169 or, h² = 169 – 25 or, h² = 144

∴ h = √144 ⇒ h = 12

∴ the height of the toy = 12 cm

Again, the volume of the toy = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 52 x 12 cc = \(\frac{2200}{7}\)cc = \(314 \frac{2}{7}\)cc

Hence the height of the toy is 12 cm and the quantity of wood required to make the toy is 314\(\frac{2}{7}\)cc.

Example 4. The quantity of iron-sheet to make a boya of right circular conical shape is 75\(\frac{3}{7}\)sq-m. If the slant height of it is 5 in, then calculate the volume of air in the boya and its height. Determine of the expenditure to colour the whole surface of the boya at the rate of ₹2.80 per square metres. [The width of the iron sheet not to be considered while calculating.]

Solution:

Given:

The quantity of iron-sheet to make a boya of right circular conical shape is 75\(\frac{3}{7}\)sq-m. If the slant height of it is 5 in, then calculate the volume of air in the boya and its height.

The slant height of the boya is 5 metres.

Let the radius, of the boya be r metres.

∴ the total surface area of the boya = \(\frac{22}{7}\)r (5+r) sq-m

∴ \(\frac{22}{7}\)r (5+r) = 75\(\frac{3}{7}\)

or, \(\frac{22}{7}\)r(5 + r) = \(\frac{528}{7}\)

or, r (r+5) = \(\frac{528 \times 7}{7 \times 22}\) or, r(r+5)= 24

or, r² + 5r- 24 = 0 or, r² + 8r- 3r- 24 = 0 or, r (r + 8) – 3 (r + 8) = 0 or, (r + 8) (r – 3) = 0

∴ either r + 8 = 0 or, r – 3 = 0

⇒ r = -8  ⇒ r = 3

Since the radius can never be negative, ∴ r = 3.

Now, if the height of the boya be h m, then h² + 3² = 5² or, h² = 25-9 or, h² = 16 or, h = 4.

∴ the height of the boya is 4 metres.

∴ the volume of boya = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 3² x 4 cubic-metres

= \(\frac{264}{7}\) cubic-metres = 37\(\frac{5}{7}\) cubic-metres

Hence the height of the boya = 4 metres and the volume of air in. the boya =37\(\frac{5}{7}\) cubic- metres.

Also, at the rate of ₹2.80 per square ‘metre, the cost of colouring the whole surface of the boya

= ₹2.80 x \(\frac{528}{7}\) = ₹211.20

Example 5. In a right circular conical tent 11 persons can stay, For each person 4 sq-m space in the base and 20 m3 air are necessary. Determine the height of the tent put up exactly for 11 persons.

Solution:

Given:

In a right circular conical tent 11 persons can stay, For each person 4 sq-m space in the base and 20 m3 air are necessary.

Let the radius of the base of the tent be r metre.

∴ the area of the base = πr² sq-metre

Again, space required for staying for 11 persons =11×4 sq-m. = 44 sq-m.

∴ \(\frac{22}{7}\) x r² =44 ⇒ r² =14.

Again, the total air required for 11 person = 20 x 11 m³ = 220 m³.

Now, if the height of the.terit be h metre, then its volume = \(\frac{1}{3}\)πr²h cubic- metres.

As per question, \(\frac{1}{3}\) x \(\frac{22}{7}\) x r²h = 220.

or, \(\frac{1}{3}\) x \(\frac{22}{7}\) x 14 x h = 220 [∵ r² = 14]

or, h = \(\frac{220 \times 3 \times 7}{22 \times 14}\) or, h = 15

Hence the height of the tent = 15 metres.

Example 6. The external diameter of a conical-coronet made off thermocol is 21 cm in length. To wrap up the outer surface of the coronet with foil, the expenditure will be ₹57.75 at the rate of 10p per-m². Calculate the height and slant height of the coronet.

Solution:

Given:

The external diameter of a conical-coronet made off thermocol is 21 cm in length. To wrap up the outer surface of the coronet with foil, the expenditure will be ₹57.75 at the rate of 10p per-m².

The external diameter of the coronet = 21 cm.

∴ the external diameter of the coronet = \(\frac{21}{2}\)

Let the slant height of the coronet be 1 cm,

∴ the curved surface area = \(\frac{22}{7}\) x \(\frac{21}{2}\) x l sq-cm = 33 l sq-cm

10 paise = ₹0.1 is spent to wrap up 1 sq-cm

∴ ₹57.75 is spent to wrap up \(\frac{1 \times 57.75}{0 \cdot 1}\) sq -cm.= 577.5 sq-cm.

As per question, 33 l = 577.5 ⇒ l = \(\frac{577 \cdot 5}{33}\) = 17.5

Let the height of the coronet be h cm.

∴ h² + \(\left(\frac{21}{2}\right)^2\) = (17.5)2 or, h² + 110.25 = 306.25 or, h² = 306.25 – 110.25

or, h² = 196 or, h = 14.

Hence the height of the coronet is 14 cm and the slant height is 17.5 cm.

Example 7. The base area of a right circular cone is 21 m and height is 14 m. Calculate the expenditure to colour the curved surface at the rate of ₹1.50 per sq-m.

Solution:

Given:

The base area of a right circular cone is 21 m and height is 14 m.

The diameter of the base of the cone = 21 metres

∴ the radius of the base of the cone = \(\frac{21}{2}\) metres

The height of the cone = 14 metres

∴ the slant height of the cone = \(\sqrt{(14)^2+\left(\frac{21}{2}\right)^2}\) metres = \(\sqrt{196+\frac{441}{4}}\) metres

= \(\sqrt{\frac{784+441}{4}}\) metres = \(\sqrt{\frac{1225}{4}}\) metres = \(\frac{35}{2}\) metres.

∴ the curved surface area = \(\frac{22}{7} \times \frac{21}{2} \times \frac{35}{2}\)sq- metres = 577.5 sq-m.

So, the expenditure to colour the curved surface area at the rate of ₹1.50 per sq-m = ₹577.5 x 1.50 = ₹866.25

Hence the required cost = ₹866.25

Example 8. The shape of a heap of wheat is a right circular cone, the diameter of base of which is 9 metres and height is 3.5 metres. Find the volume of the heap of wheat. How much sq-metre of plastic sheet will be needed to cover that heap of wheat? [Given that n = 3.14 and 130 = 11.4]

Solution:

Given:

The shape of a heap of wheat is a right circular cone, the diameter of base of which is 9 metres and height is 3.5 metres.

The radius of the heap of wheat = \( metres

Height = 3.5 metres.

∴ the volume of the heap of wheat = [latex]\frac{1}{3}\) x π x \(\left(\frac{9}{2}\right)^2\) x 3.5 cubic- metres

= \(\frac{1}{3}\) x 3.14 x (4.5)² x 3.5 cubic-metres = 74.18 cubic-metres

If the slant height of the heap of wheat be l metre, then

\(l^2=(3 \cdot 5)^2+\left(\frac{9}{2}\right)^2=\left(\frac{7}{2}\right)^2+\left(\frac{9}{2}\right)^2=\frac{49}{4}+\frac{81}{4}=\frac{130}{4}\)

∴ l = \(\sqrt{\frac{130}{4}}\) = 5.7 (approx.)

∴the curved surface area of the heap of wheat

= π x \(\frac{9}{2}\) x 5.7sq-m= 3.14 x 4.5 x 5.7 sq-m = 80.54 sq-m.

Hence the total volume of the heap of wheat is 7418 cubic metres and to cover the heap 80.54 sq-m of the plastic sheet will be needed.

Example 9. What will be the height of the cone of radius 12 cm if a solid cone is made by melting a solid cone of radius of base 6 cm and of slant height 10 cm?

Solution: The radius of the base of the first right circular cone = 6 cm and slant height = 10 cm.

∴ the height of the right circular cone = \(\sqrt{(10)^2-(6)^2}\) cm = 8cm

∴ volume of the cone = \(\frac{1}{3}\)π x 6² x 8 cu-cm

The new radius of the cone is 12 cm and the height is h cm, the volume of the cone = \(\frac{1}{3}\)π x (12)² x h cubic- cm.

As per question, \(\frac{1}{3}\)π x(12)² xh = \(\frac{1}{3}\)π x (6)² x 8

⇒ h = \(\frac{6 \times 6 \times 8}{12 \times 12}\) = 2.

Hence the height of the new cone = 2 cm.

Example 10. What changes will be occured in the volume of a right circular cone if its radius be increased by 10% and its height be decreased by 10%?

Solution: Let the radius of the right circular cone be r units and height be h units.

∴ the volume of the cone = \(\frac{1}{3}\)πr²h cubic-units

Now, if the radius be Increased by 10%, then the radius become (r + r x \(\frac{10}{100}\))units = \(\frac{11r}{10}\)units

Again, if the height of the cone be decreased by 10%, then its height becomes

\(\left(h-h \times \frac{10}{100}\right)\) units = \(\frac{9h}{10}\)units.

As a result of these increasing and decreasing, the volume of the cone becomes

\(\pi \times\left(\frac{11 r}{10}\right)^2 \times \frac{9 h}{10}\)cu – units = \(\pi \times \frac{121 r^2}{100} \times \frac{9 h}{10} \)cubic-units = \(\frac{1089}{1000} \pi r^2 h\)cubic -units

∴ the increment in volume = \(=\left(\frac{1089}{1000} \pi r^2 h-\frac{1}{3} \cdot \pi r^2 h\right)\)cu-units = \(\frac{2267}{3000}\)πr²h cubic -units.

So, the percentage of increment in volume = \(=\frac{\frac{2267}{3000} \pi r^2 h}{\pi r^2 h} \times 100 \%=\frac{2267}{3000} \times 100 \%\) = 75.57%

Hence the volume of the cone will be increased by 75.57 % (approx.)

“WBBSE Class 10 Maths mensuration solved problems on cones”

Example 11. If the whole surface area of a right circular cone be 3696 sq-cm and the ratio of its radius of base and height be 3: 4, then what will be the curved surface area of the cone?

Solution:

Given:

If the whole surface area of a right circular cone be 3696 sq-cm and the ratio of its radius of base and height be 3: 4,

The ratio of the radius of base and height of the cone is 3: 4; then let the radius of the cone be 3x cm and height be 4x cm.

So, the slant height of the. cone = \(\sqrt{(3 x)^2+(4 x)^2}\) cm = \(\sqrt{25 x^2}\)cm = 5x cm.

The whole surface area of the cone = 3696 sq-cm

∴ \(\frac{22}{7}\) x 3x x(3x + 5x) = 3696

or, \(\frac{22}{7}\) x 3x x 8x = 3696

or, \(x^2=\frac{3696 \times 7}{22 \times 3 \times 8}\)

or, x² = 49 or, x = √49 = 7.

So, the radius of base of the cone = 3x cm = 3 x 7 cm = 21 cm

and slant height of the cone = 5x cm = 5 x 7 cm = 35 cm

∴ Curved surface area of the cone = \(\frac{22}{7}\) x 21 x 35 sq-cm = 2310 sq-cm

Hence the required curved surface of the right circular cone = 2310 sq-cm.

Example 12. If the curved surface area of a toy of shaped a right circular cone of height 24 cm is 550 sq-cm, then what will be the volume of the toy?

Solution:

Given:

If the curved surface area of a toy of shaped a right circular cone of height 24 cm is 550 sq-cm

Let the radius of base of the toy be r cm and its slant height be l cm.

Then, l² = r² + 24² [∵ height = 24 cm] ⇒ l² = r² + 576 ⇒ l = \(\sqrt{r^2+576}\).

The curved surface area of the toy = 550 sq-cm.

∴ \(\frac{22}{7}\) x r x l = 550

or, r x \(\sqrt{r^2+576}\) = \(\frac{550 \times 7}{22}\)

or, r x \(\sqrt{r^2+576}\) = 25 x 7

or, r² x (r² + 576) = (25 x 7)² or, r⁴ + 576 r² = 625 x 49 or, r⁴ + 576r² – 625 x 49 = 0

or, r⁴ + (625 – 49)r² – 625 x 49 = 0 or, r⁴ + 625r² – 49 r² – 625 x 49 = 0

or, r² (r² + 625) – 49 (r² + 625) = 0 or, (r² + 625) (r² – 49) = 0

∴ either r² + 625 = 0, which is impossible, or, r² – 49 = 0 ⇒ r² = 7² ⇒ r = 7.

So, the radius of the base of the toy = 7

∴ the volume of the toy = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 7² x 24 cubic-cm

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x 49 x 24 cc = 1232 cc

Hence the required volume of the toy = 1232 cc.

Example 13. The volume of a right circular cone is 410\(\frac{2}{3}\)cc. If the radius of the base of the cone be 14 cm, then what will be the curved surface area of the cone?

Solution:

Given:

The volume of a right circular cone is 410\(\frac{2}{3}\)cc. If the radius of the base of the cone be 14 cm

Let the height of the cone be h cm.

The radius of the cone = 14 cm,

∴ the volume of the cone = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 14² x h cc.

As per question, \(\frac{1}{3}\) x \(\frac{22}{7}\) x 14² x h = 410\(\frac{2}{3}\)

or, \(\frac{22}{3}\) x 28 h = \(\frac{1232}{3}\)

or, h = \(\frac{1232 \times 3}{3 \times 22 \times 28}\) or, h = 2

∴ the height of the cone = 2 cm.

∴ the slant height of the cone = \(\sqrt{(14)^2+(2)^2}\)cm = 200 cm = \(\sqrt{100 \times 2}\) cm = 10√2 cm

∴ the curved surface area of-the cone = \(\frac{22}{7}\) x 14 x 10V2 sq- cm = 440√2 sq- cm

Hence the required curved surface of the cone = 440√2 sq-cm

Example 14. What changes will be occured in the height of a right circular cone if its radius of base is decreased by 50% and its volume be decreased by 25%?

Solution: Let the radius of base of the cone be r units and height be h units.

Now, if the volume of the cone be V cubic-units, then V = \(\frac{1}{3}\)πr²h…..(1)

Now, let the volume of the cone be V₁ cubic-unit and height be h₁ units when its radius of base is decreased by 50% and volume is decreased by 25%.

∴ \(\quad \mathrm{V}_1=\frac{1}{3} \pi\left(r-r \times \frac{50}{100}\right)^2 \times h_1=\frac{1}{3} \pi\left(r-\frac{r}{2}\right)^2 \times h_1=\frac{1}{3} \pi \times \frac{r^2}{4} \times h_1\)

i.e, \(\mathrm{V}-\mathrm{V} \times \frac{25}{100}=\frac{1}{3} \pi \times \frac{r^2}{4} \times h_1\)

or, \(\frac{3 \mathrm{~V}}{4}=\frac{1}{3} \pi \times \frac{r^2}{4} \times h_1\)

or, \(9 \mathrm{~V}=\pi r^2 h_1\)

or, \(9 \times \frac{1}{3} \pi r^2 \dot{h}=\pi r^2 h_1\) [by (1)]

or, \(3 h=h_1\)

∴h₁=3h

∴ the percentage of change in height = \(\frac{3 h-h}{h}\) x 100% = \(\frac{2h}{h}\) x 100% =200%

Hence the required increase in height in percent of the cone = 200%.

Example 15. What changes in curved surface area of a right circular cone will be occured if its radius of base and height is doubled?

Solution:

Let the radius of base, height and slant height of the right circular cone be r unit, h unit and l unit respectively.

∴ l = \(\sqrt{h^2+r^2}\)…….(1)

Now if the curved surface area of the cone be A sq-unit, then

A = πrl……(2)

Also, if the radius of base and the height of the cone be doubled, then let the slant height of it be l₁ units

∴ \(l_1=\sqrt{(2 r)^2+(2 h)^2}=\sqrt{4 r^2+4 h^2}=\sqrt{4\left(r^2+h^2\right)}=2 \sqrt{h^2+r^2}\) = 2l [by (1)]…..(3)

In this case, if the curved surface area be A₁ sq-unit, then

A₁ = π x 2r x lπ = π x 2r x 2l [by (3)] = 4πrl = 4A [by (2)]

Hence if the radius of base and height of a right circular cone be doubled, then the curved surface area becomes 4 times of its previous curved surface area.

Example 16. If the curved surface area, volume, height and semi-vertical angle of a right circular cone be S,V.h and respectively, then prove that S= \(\frac{\pi h^2 \sin a}{\cos ^2 a}\) and V = \(\frac{1}{3} \pi h^3 \tan ^2 \alpha\)

Solution:

Given:

If the curved surface area, volume, height and semi-vertical angle of a right circular cone be S,V.h and respectively,

Here,, the height, AB = h, ∠BAC = α = semi-vertical angle, radius of base, BC = r and slant height, AC = l.

∴ \(\frac{r}{h}\) = tan α or, h tan α= r……(1)

and \(\frac{h}{l}\) = cosα or, l = \(\frac{h}{\cos \alpha}\)……(2)

Now, the curved surface are of the cone = πrl.

= π x h tan α x \(\frac{h}{\cos \alpha}\) [by (1) and (2)]

= x h.\(\frac{\sin x}{\cos \alpha} \times \frac{h}{\cos \alpha}=\frac{\pi h^2 \sin \alpha}{\cos ^2 \alpha}\)

∴ S = \(\frac{\pi h^2 \sin \alpha}{\cos ^2 \alpha}\)

Moreover, the volume of the cone, V = \(\frac{1}{3}\)πr²h = \(\frac{1}{3}\) (h tan α)².h [by (1)]

= \(\frac{1}{3} \pi \cdot h^2 \tan ^2 \alpha \cdot h=\frac{1}{3} \pi h^3 \tan ^2 \alpha\)

Hence V = \(\frac{1}{3} \pi \cdot h^3 \tan ^2 \alpha\)

“Curved surface area of right circular cone examples Class 10”

Example 17. If the height, curved surface area and volume of a right circular cone be A, c and v respectively, then prove that 3πvh³ – c²h + 9v² = 0.

Solution:

Given:

If the height, curved surface area and volume of a right circular cone be A, c and v respectively

Let the radius of base and slant height of the cone be r unit and l unit respectively.

∴ l² = h² + r²……(1)

Again, v = \(\frac{1}{3}\)πr²h ……(2) and c = πrl…….(3)

∴ 3πvh³ – c²h² + 9V²

= 3π x \(\frac{1}{3}\)πr²h x h³ -(πrl)² x h² +9 x (\(\frac{1}{3}\)πr²h)² [by (2) and (3)]

= \(\pi^2 r^2 h^4-\pi^2 r^2 l^2 h^2+\pi^2 r^4 h^2\)

= \(\pi^2 r^2 h^4-\pi^2 r^2\left(r^2+h^2\right) \cdot h^2+\pi^2 r^4 h^2\) [by (1)]

= \(\pi^2 r^2 h^4-\pi^2 r^4 h^2-\pi^2 r^2 h^4+\pi^2 r^4 h^2\) = 0 .

Hence 3πvh³ – c²h² + 9v² = 0 [Proved]

Example 18. The height of a right circular cone is 30 cm. A small part of the cone is cut off from the above part of the cone with the help of a plane parallel to the base of the cone. If the volume of the small part be 1/27 part of the volume of the original cone, then at what distance above the base of the cone, it has been cut off?

Solution:

Given:

The height of a right circular cone is 30 cm. A small part of the cone is cut off from the above part of the cone with the help of a plane parallel to the base of the cone. If the volume of the small part be 1/27 part of the volume of the original cone

Let OAB is the original cone of radius of base R cm and its height be H cm.

∴ the volume of the cone = \(\frac{1}{3}\)πR²Hcc = \(\frac{1}{3}\)πR² X 30CC = 10πR²cc

The cut-off small part of the cone is OCD, the radius of base of which is r cm and height is h cm.

Then the volume of this part = \(\frac{1}{3}\)πr²h cc.

As per question, \(\frac{1}{3}\)πr²h = \(\frac{1}{27}\)(10πR²)

⇒ h = \(\frac{10 \pi \mathrm{R}^2}{27} \times \frac{3}{\pi r^2}\)

h= \(\frac{10}{9} \cdot \frac{\mathrm{R}^2}{r^2}\)…….(1)

Now, from ΔOQB and ΔOPD we get, \(\frac{\mathrm{QB}}{\mathrm{PD}}=\frac{\mathrm{OQ}}{\mathrm{OP}}=\frac{30}{h} \Rightarrow \frac{\mathrm{R}}{r}=\frac{30}{h}\)

∴ from (1) we get, h = \(\frac{10}{9} \cdot\left(\frac{\mathrm{R}}{r}\right)^2=\frac{10}{9} \cdot\left(\frac{30}{h}\right)^2=\frac{10}{9} \times \frac{900}{h^2}\)

or, h³ = 1000 or h³ = (10)³ or, h = 10

∴ the small part is cut off at a distance of (30- 10) cm = 20 cm high above the base of the cone.

Hence the required height = 20 cm.

Example 19. The lengths of the adjacent sides of right angle of a right-angled triangle are 20 cm and 15 cm. What will be the total volume of the two right circular cones when this triangle is rotated with its hypotenuse taken as the axis?

Solution:

Given:

The lengths of the adjacent sides of right angle of a right-angled triangle are 20 cm and 15 cm.

Let the perpendicular of the right-angled triangle ABC be ABC = 20 cm and its base, AB = 15 cm.

∴ hypotenuse, AC = \(\sqrt{(20)^2+(15)^2} \mathrm{~cm}\)

= \(\sqrt{400+225} \mathrm{~cm}=\sqrt{625} \mathrm{~cm}=25 \mathrm{~cm}\)

Now, perpendicular BO is drawn from B to the hypotenuse AC.

∴ the area of the ΔABC = \(\frac{1}{2}\) x AB x BC = \(\frac{1}{2}\) x AC x BO

⇒ AC x BO = AB x BC ⇒ 25 x BO = 15 x 20

⇒ BO = \(\frac{15 \times 20}{25}\) => BO = 12

So, the radius of two cones produced by rotating the triangle with the hypotenuse AC as the axis will be 12 cm.

Also, the heights of the two cones will be AO and CO.

So, the total volume of the two cones = \(\left(\frac{1}{3} \pi \times 12^2 \times \mathrm{AO}+\frac{1}{3} \pi \times 12^2 \times \mathrm{CO}\right)\)cc

= \(\frac{1}{3}\)π x (12)² x (AO + BO)cc = \(\frac{1}{3}\)π x 144 x AC cc

= \(\frac{1}{3}\) x 144 x 25π cc = 1200π cc

Hence the total volume of the two produced cones = 1200π cc.