Revision: Mensuration Geometry Maths 2 SSC (English Medium) 10th Standard Maharashtra State Board

Sector of a circle
Region enclosed by two radii and the corresponding arc.

Minor sector
Sector with angle < 180°.

Major sector
Sector with angle > 180°.
Angle of major sector = 360° − angle of minor sector.

Segment of a circle
Region enclosed by a chord and the corresponding arc.

Minor segment
A smaller region formed by the chord.

Major segment
The remaining larger region of the circle.

\[\text{Area of sector}=\frac{\theta}{360}\times\pi r^2\]

\[\text{Length of arc}=\frac{\theta}{360}\times2\pi r\]

Area of minor segment = Area of sector − Area of triangle

\[A(\text{minor segment})=\frac{\theta}{360}\pi r^2-\frac{1}{2}r^2\sin\theta\]

$Area\; of\; major\; segment=Area\; of\; circle-Area\; of\; minor\; segment$

That is,

A(major segment) = πr² − A(minor segment)

Let radii r₁ > r₂​, height h, slant height l

• Slant height: l = \[\sqrt{h^2+(r_1-r_2)^2}\]​

• Curved surface area of a frustum: πl(r₁ + r₂)

• Total surface area of a frustum: πl(r₁ + r₂) + πr₁² + πr₂² ​

• Volume: \[\frac{1}{3}\]πh(r₁² + r₂² + r₁ × r₂)

• Curved surface area = 2πr²

• Total surface area of a solid hemisphere = 3πr²

• Volume = \[\frac{2}{3}\]πr³

• Slant height l =\[\sqrt{h^2+r^2}\]​

• Curved surface area = πrl

• Total surface area = πr(r + l)

• Volume = \[\frac{1}{3}\] × πr²h

Cube (side = a)

• Lateral surface area = 4a²

• Total surface area = 6a²

• Volume = a³

• Lateral surface area = 2h(l + b)

• Total surface area = 2(lb + bh + hl)

• Volume = lbh

• Curved surface area = 2πrh

• Total surface area = 2πr(r + h)

• Volume = πr²h

• Surface area = 4πr²

• Volume = \[\frac{4}{3}\]πr³

Given angle subtended at centre of circle = 𝜃

∠OAB = 90° [At joint of contact, tangent is perpendicular to radius]

OAB is right angle triangle

Cos 𝜃 =`(adj.side)/(hypotenuse) =r/OB`⇒ 𝑂𝐵 = 𝑟 sec 𝜃 … … (𝑖)

tan 𝜃 =`(opp.side)/(adju.side)=AB/r`⇒ 𝐴𝐵 = 𝑟 tan 𝜃 … … . (𝑖𝑖)

Perimeter of shaded region = AB + BC + (CA arc)

= 𝑟 tan 𝜃 + (𝑂𝐵 − 𝑂𝐶) +`theta/360^@`× 2𝜋𝑟

= 𝑟 tan 𝜃 + 𝑟 sec 𝜃 − 𝑟 +`(pithetar)/180^@`

= 𝑟 (tan 𝜃 + sec 𝜃 +`(pitheta)/180^@`− 1)

Given angle subtended at centre of circle = 𝜃

∠OAB = 90° [At joint of contact, tangent is perpendicular to radius]

OAB is right angle triangle

Cos 𝜃 =`(adj.side)/(hypotenuse) =r/OB`⇒ 𝑂𝐵 = 𝑟 sec 𝜃 … … (𝑖)

tan 𝜃 =`(opp.side)/(adju.side)=AB/r`⇒ 𝐴𝐵 = 𝑟 tan 𝜃 … … . (𝑖𝑖)

Area of shaded region = (area of triangle) – (area of sector)

`= (1/2× OA × AB) −theta/360^@× pir^2`

`=1/2× r × r tan theta −r^2/2[theta/180^@× pi]`

=`r^2/2[tantheta −(pitheta)/180^@]`

Definition: A segment is a region of a circle bounded by a chord and its arc

Two Main Types:

• Minor Segment = smaller piece

• Major Segment = larger piece

Semicircle Special Case:

• Formed when chord = diameter

• Creates two perfectly equal segments

• Each semicircle = half the circle's area