Krishna
0

Step 1; Note down the given values and examine the given figure

Step 2: Find the unknown area by the known areas.

NOTE: Area of the segment AYB = Area of sector OAYB − Area of ∆OAB

Step 3: Calculate the area of the sector

[Step 1: Recall the area of the sector formula

NOTE: \frac{\theta}{360} * \pi r^2

Step 2: Substitute all the values in the formula.

EXAMPLE:   \frac{\theta}{360} * \pi r^2

\frac{120}{360}\ *\ 3.14\ *\ 21^2

Step 3: Simplify the equation

EXAMPLE:  \frac{1}{3}\cdot\ 3.14\ \cdot\ 441\ cm^2

462\ cm^2]

Step 4: Find the area of the triangle

Skill 1: Recall the area of the circle formula

NOTE: \frac{1}{2} base * height

Skill 2: Construct a line joining the center and the mid point of the chord and

note down the changes

NOTE: draw  (O center, M mid point)OM ⊥ AB(Chord) as shown in

the figure:- Note OA = OB.  Therefore, by RHS congruence,∆AMO ≅ ∆BMO

So, M is the midpoint of AB and ∠ AOM = ∠ BOM = \frac{1}{2}120 = 60 \degree

Skill 3: Find the height using the trigonometric ratios.

EXAMPLE:  Let, OM = x cm

So, from ∆OMA ,   \cos 60 = \frac{OM}{OA}

\frac{1}{2} = \frac{x}{21} (since   \cos 60 = \frac{1}{2}

OM = x = \frac{21}{2}

Also,

\sin 60 = \frac{AM}{OA}

\frac{\sqrt{3}}{2} = \frac{AM}{21}  (since   \sin 60 = \frac{\sqrt{3}}{2}

AM = \frac{21\sqrt{3}}{2}

.            Skill 4: Calculate the base of the triangle

NOTE: Therefore AB = 2AM

EXAMPLE: AB = 2 * \frac{21\sqrt{3}}{2}

AB = 21 * \sqrt{3}

Skill 5: Substitute the calculated values in the formula

EXAMPLE: \frac{1}{2} * AB * OM

\frac{1}{2} *(21 * \sqrt{3}) (\frac{21}{2})

\frac{441}{4} \sqrt{3}

Step 5: Determine the area of the segment

Area of the segment AYB = Area of sector OAYB − Area of ∆OAB

EXAMPLE: Area of the segment AYB = 462 -   \frac{441}{4} \sqrt{3}

Area of the segment AYB = 271.047 cm^2