Question

Find the area of the segment AYB shown in fig.12.9 , if the radius of the circle is 21 cm and angle 120° . { use π =22/7}

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Answer 1

Step-by-step explanation:

• SEGMENT - AREA BOUNDED BY AN ARC AND A CHORD

• SECTOR - AREA BOUNDED BY AN ARC AND TWO RADII

• RADII - PLURAL OF RADIUS

• AREA OF SECTOR - Ѳ/360° X πr²

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AREA OF SEGMENT AYB = AREA OF SECTOR OAYB - AREA OF ∆OAB

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AREA OF SECTOR = Ѳ/360° X πr²

AREA OF SECTOR OAYB = 120°/360° X 22/7 X 21 X 21 cm²

AREA OF SECTOR OAYB = 462cm²

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AREA OF ∆OAB

We draw a OM ⊥ AB in ∆AOB.

Now in ∆ AMO and ∆ BMO.

We have

<AMO = <BMO = 90°

AO = BO (RADIUS) [HYPOTENUSE]

MO = MO (COMMON)

.•. ∆ AMO ≅ ∆ BMO. [BY RHS CONGRUENCE CRITERION]

AM = BM [BY CPCT]

THEREFORE , M IS THE MIDPOINT OF LINE AM.

∠ AOM = ∠ BOM [ BY CPCT]

<AOM = <BOM = 1/2 X 60°

THEREFORE <AOM = <BOM = 60°.

NOTE : YOU NEED TO LEARN TRIGONOMETRY BEFORE SOLVING THIS QUESTION.

USING TRIGONOMETRY NOW.

In ∆OMA

When <AOM is Ѳ.

Then ,

AM = PERPENDICULAR

OM = BASE

OA = HYPOTENUSE

Let OM = x cm

cosѲ = BASE/HYPOTENUS

cos 60° = 1/2

cos60° = OM/OA

1/2 = x/21

→ x = 21/2cm

OM = 21/2cm

sinѲ = PERPENDICULAR/HYPOTENUSE

sin60° = √3/2

sin 60° = AM/OA

√3/2 = AM/21

→ AM = 21√3/2 cm

AB = 2 X AM

AB = 2 X 21√3/2

AB = 21√3 cm

AREA OF ∆ = 1/2 X BASE X ⊥ HEIGHT

AREA OF ∆ OAB = 1/2 X AB X OM

AREA OF ∆ OAB = 1/2 X 21√3 X 21/2 cm²

AREA OF ∆OAB = 441/4 √3 cm²

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AREA OF SEGMENT AYB = AREA OF SECTOR OAYB - AREA OF ∆OAB

AREA OF SEGMENT AYB = (462 - 441/4√3)cm²

AREA OF SEGMENT AYB = 21/4 (88 - 21√3) cm²

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HOPE IT HELPS YOU

THANKS !

Answer 2

Answer :

Area of AYB = 270. 86 cm²

Formulae applied :

• Area of sector = x/360 × πr²
• Area of triangle = ½ bh

Explanation :

From the figure,

Area of AYB = Area of sector (AOBYA) - Area of ∆AOB

Area of sector (AOBYA) = x/360 × πr²

= 120/360 × 22/7 × 21 × 21

= 462 cm²

In triangle AOB,

AO = BO

i.e, $\angle$ OAB = $\angle$ OBA

→ 120 + $\angle$ OAB + $\angle$ OBA = 180°

→ 120 + 2($\angle$ OAB) = 180

→ $\angle$ OAB = 30° = $\angle$ OBA

In ∆ODA, sin 30° = OD/21

OD/21 = 1/2

OD = 10.5 cm

cos 30° = AD/21

AD/21 = √3/2

AD = 10.5√3

AB = 2(10.5√3)

AB = 21√3

Thus, base of triangle, AB = 21√3 cm

Height of triangle = 10.5 cm

Area of triangle = ½bh = ½ (21√3)(10.5)

= 190.14 cm²

→ Area of AYB = Area of sector (AOBYA) - Area of ∆AOB

→ Area of AYB = 462 - 190.14

$\boxed{Area\:\: of \:\:AYB = 270.86 cm²}$