Question

Find the area of the segment AYB shown in the figure, if the radius of the circle is 21 cm and ∠ AOB = 120°

Answer 1

Answer:

$\huge{\boxed{\mathscr\pink{♡}{\mathscr{answer}}\pink{♡}}}$

In ∆AOB, draw a perpendicular line from O which intersect AB at M.

➡In ∆AOM, angle AMO = 90

➡angle OAM = 30

cos 30 = AM/AO

√3/2 = AM/21

AM = 21×√3/2

AB = 2(AM)

=2(21×√3/2)

=21√3

➡OM^2 = AO^2-AM^2

=21^2-(21√3/2)^2

=441-330.51

=110.48

OM =√110.48

OM =10.51

➡OM = 10.51cm

➡Area of ∆AOM = 1/2 AB × OM

=1/2 ×21√3 ×10.51

=191.14cm^2

➡Area of sector AOBY = 120πr^2/360

=120×21×21×22/2520

=462cm^2

➡Area of segment AYB = Area of sector OAYB -Area of∆OAB

=462-191.14

=270.86

➡Area of segment AYB is 270.86cm^2.

Answer 2

Given

• Radius of circle(r) = 21cm
• ∠AOB = 120°
• OA = OB (radius)

To Find

• Area of segment AYB.

Solution

Area of the segment AYB = (Area of sector OAYB) - (Area of ∆OAB)...............eq.(1)

Now,

Area of sector OAYB,

$\tt = \frac{ \theta}{360} \times \pi \times r {}^{2} \\ \\ \tt \: = \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \\ \\ \tt= \frac{12}{36} \times \frac{22}{7} \times 441 \\ \\ \tt = \frac{252}{252} \times 441 \\ \\ \tt = 462cm {}^{2}.........eq.(2)$

✏ Draw OC ⊥ AB.

____________________________

⇒In ∆ AMO and ∆ BMO,

OA = OB _______(radius)

OC = AC _______(common)

∠OCA = ∠OCB_______ (OC ⊥ AB)

∵ ∆ AMO ≅ ∆ BMO ( By RHS congruence )

∠AOB = 120°

Then,

∠AOC = ∠BOC = 120/2 = 60°

⇒In ∆OCA,

Perpendicular = OC

Hypotenuse = OA

$\tt \implies \: cos \theta = \frac{perpendicular}{hypotenuse} \\ \\ \tt \implies cos \: 60 = \frac{AC}{OA} \\ \\ \tt \implies \: \frac{1}{2} = \frac{OC}{21} \\ \\ \tt \implies \: OC = \frac{21}{2} cm$

Similarly,

$\tt \implies \: sin \: 60 = \frac{AC}{OA} \\ \\ \tt \implies \frac{ \sqrt{3} }{2} = \frac{AC}{21} \\ \\ \tt \implies \: AC = \frac{21 \sqrt{3} }{2}$

★AB = 2AM

AB = 2(21√3/2)

AB = 21√3 cm

Now,

$\tt \implies \: area \: of \: triangle \: AOB = \frac{1}{2} \times AB \times OC \\ \\ \tt \implies \: \frac{1}{2} \times 21 \sqrt{3} \times \frac{21}{2} \\ \\ \tt \implies (\frac{441}{4} ) \sqrt{3} cm {}^{2}.......eq.(3)$

From equation 1, 2 and 3.

$\tt \implies \: Area \: of \: the \: segment \: AYB \\ \\ = 462 - (\frac{441}{4} \times \sqrt{3} )$

$\large \implies \boxed {\boxed {\tt \blue {\frac{21}{4} (81 - 21 \sqrt{3} )cm {}^{2} }}}$