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Mensuration

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CAT 2025 Lesson : Mensuration - Cone, Pyramid & Frustum

3.5 Right Circular Cone

A right circular cone is a circular cone whose tip (or vertex) is perpendicular to the centre of its circular base. This is very often referred to as just cone in the entrance tests.

In the adjacent figure, $\bm{r}$ is the radius of the base. $\bm{h}$ is the height of the cone. $\bm{l}$ is the slant height of the cone, wherein $l = \sqrt{r^{2} + h^{2}}$ Volume of cone = $\dfrac{1}{3}\pi r^{2}h$ Curved Surface Area (CSA) = $\pi rl$ Flat Surface Area (FSA) = $\pi r^{2}$ Total Surface Area (TSA) = $\pi r(l+r)$

Example 11

What is the curved surface area of a cone (in cm

^2

) whose radius and height are in the ratio of 3 : 4 and whose volume is 324

\pi

^3

?

(1) 135

\pi

(2)150

\pi

(3) 160

\pi

(4) 180

\pi

Solution

Let the radius and height of the cone be

3x

and

4x

respectively.

Volume of the cone =

\dfrac{1}{3}\pi r^{2}h

324\pi

⇒

\dfrac{1}{3} \times 9x^{2} \times 4x

324

⇒

x^{3}

27

⇒

x

= 3

∴ Radius and Height are 9 cm and 12 cm respectively.
Slant Height =

l

\sqrt{r^2+ h^2}

\sqrt{9^2+ 12^2}

= 15

CSA =

\pi r l

\pi \times 9 \times 15

= 135

\pi

Answer: (1) 135

\pi

Example 12

A solid cone, with a height of 28 cm and a circular base of diameter 12 cm, is melted and made into a solid cuboid whose dimensions are in the ratio 11 : 6 : 2. What is the total surface area (in cm

^2

) of this resultant cuboid?

Solution

Volume of the cone =

\dfrac{1}{3} \pi r^2 h

\dfrac{1}{3} \times \dfrac{22}{7} \times 6^2 \times 28 = 22 \times 12 \times 4 cm^3

As their ratio is 6 : 3 : 2, let length (

l

), breadth (

b

) and height (

h

) of the cuboid be 6

x

, 3

x

and 2

x

respectively.

Volume of the cuboid =

lbh

11x \times 6x \times 2x

= 22 \times 12 \times 4 cm^3

⇒

x^3

= 8 ⇒

\bm{x}

= 2

∴

l

= 22 cm,

b

= 12 cm and

h

= 4 cm
TSA of the cuboid = 2

(lb + bh +hl)

= 2(264 + 48 + 88) = 800 cm

^2

Answer: 800

3.6 Right Pyramid

Like Right Circular Cones, all the points of a polygonal base are connected to a single tip that in a pyramid. bottom and top of a right prism are identical. The difference between the two is that the circular cone has a circular base while a pyramid has a polygonal base.

By definition, a right pyramid is a figure with a tip (or vertex) which is perpendicular to the centre of its polygonal base.

In the adjacent figure, $\bm{h}$ is the height of the right pyramid. $\bm{l}$ is the slant height of the right pyramid. Volume of right pyramid = $\dfrac{1}{3} \times$ Area of base $\times h$ Lateral Surface Area (LSA) = $\dfrac{1}{2} \times$ Perimeter of base $\times l$ Total Surface Area (TSA) = LSA + Area of Base

Example 13

What is the total area (in m

^2

) of all the slant surfaces in a right pyramid with a regular hexagonal base, whose base area is 50

\sqrt{3}m^2

and height is 12 m?

(1) 100

\sqrt{3}

(2)130

\sqrt{3}

(3) 170

\sqrt{3}

(4) 200

\sqrt{3}

Solution

If a is the length of the side of the regular hexagon, Area of Hexagon = $6 \times \dfrac{\sqrt{3}}{4} \times a^2$ = $50\sqrt{3}$ ⇒ $a^2$ = $\dfrac{100}{3}$ ⇒ a ⇒ $\dfrac{10}{\sqrt{3}}$
In the adjoining figure, the line connecting the mid point of the side of the hexagon to the centre, i.e. DO is perpendicular to the height OA. AD is perpendicular to the base in $\triangle$ ABC. $\therefore$ We get a right angled triangle AOD. AO = Height of the pyramid = 12 m

DO = Height of the equilateral triangle in the hexagon =

\dfrac{\sqrt{3}}{2} \times a

\dfrac{\sqrt{3}}{2} \times \dfrac {10}{\sqrt{3}}

= 5 m

Applying Pythagoras theorem,

AD =

\sqrt{AO^2+ DO^2}

\sqrt{12^2 + 5^2}

= 13 m

6 identical triangles form the slant surface, where the base of these triangles is

\dfrac {10}{\sqrt{3}}

m and height of 13 m.

∴ Total slant surface area =

6 \times \dfrac {1}{2} \times \dfrac {10}{\sqrt{3}} =130\sqrt{3}

Answer: (2)

130\sqrt{3}

3.7 Frustum of a Cone

When a cut parallel to the base of a cone is made, we would get a smaller cone (that contains the tip) and a frustum of the cone that contains the circular base of the initial cone.


In the adjacent figure, $\bm{r}$ is the radius of the smaller circular top. $\bm{R}$ is the radius of the larger circular base. $\bm{h}$ is the height of the frustum $\bm{l}$ is the slant height of the frustum.wherein $l = \sqrt{(R – r)^2 + h^2}$ Volume of frustum= $\dfrac{1}{3} \pi h \times (R^2 +Rr +r^2)$ Curved Surface Area (CSA) = $\pi l(R + r)$ Flat Surface Area (FSA) = $\pi R^2 + \pi r^2$ Total Surface Area (TSA) = $\pi (lR + lr + R^2 + r^2)$

In the adjacent figure,

\bm{r}

is the radius of the smaller circular top.

\bm{R}

is the radius of the larger circular base.

\bm{h}

is the height of the frustum

\bm{l}

is the slant height of the frustum.wherein

l = \sqrt{(R – r)^2 + h^2}

Volume of frustum=

\dfrac{1}{3} \pi h \times (R^2 +Rr +r^2)

Curved Surface Area (CSA) =

\pi l(R + r)

Flat Surface Area (FSA) =

\pi R^2 + \pi r^2

Total Surface Area (TSA) =

\pi (lR + lr + R^2 + r^2)

Example 14

When a cut is made parallel to the base of a right circular cone, we get a frustum with slant height of 5 cm and curved surface area of 105

\pi cm^2

. If the radius of the base of the right circular cone was 12 cm, then what was its height (in cm)?

Solution


Let DE be the radius( $r$ ) of the frustum’s smaller circular top, BC be the radius( $R$ ) of larger circular base and AC be the slant height( $l$ ) of the frustum. Curved Surface Area of Frustum = $\pi (r + R) l$ = 105 $\pi$ ⇒ $\pi (r + 12) 5$ = 105 $\pi$ ⇒ r + 12 = 21, r = 9 In $\triangle$ ABC and $\triangle$ ADE $\angle$ ABC = $\angle$ ADE (Right Angle) $\angle$ BAC = $\angle$ DAE (Common Angle) $\triangle$ ABC ~ $\triangle$ ADE(AA Rule)

Let DE be the radius(

r

) of the frustum’s smaller circular top, BC be the radius(

R

) of larger circular base and AC be the slant height(

l

) of the frustum.

Curved Surface Area of Frustum =

\pi (r + R) l

= 105

\pi

⇒

\pi (r + 12) 5

= 105

\pi

⇒ r + 12 = 21, r = 9

In

\triangle

ABC and

\triangle

ADE

\angle

ABC =

\angle

ADE (Right Angle)

\angle

BAC =

\angle

DAE (Common Angle)

\triangle

ABC ~

\triangle

ADE(AA Rule)

Let

x

cm be the slant height of the smaller cone that was cut.

\therefore

\dfrac{AC}{BC}

\dfrac{AE}{DE}

⇒

\dfrac{x + 5}{12}

\dfrac{x}{9}

⇒

9 (x +5) = 12x

⇒

3 x +15 = 4x, x = 15

AC = 20 cm

Applying Pythagoras theorem,

AB =

\sqrt{AC^2 – BC^2}

\sqrt{20^2 – 12^2}

= 16 cm

\therefore

The height AB of the right circular cone = 16 cm

Answer: 16 cm

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