Question

a cresent is formed by two circles which touch at A . C is the centre of the larger circle . the width of the crescent at BD is 9 CM and at EF it is 5cm. find (1) the radii of two circles (2) the area of shaded region

please answer both parts

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Answer 1

Answer:

Let the radii of the larger and smaller circles be R and r respectively. Then,

BD=9cm

⇒ 2R−2r=9

⇒ R−r=4.5

Join AE and DE. Let ∠CAE=θ Then, ∠AEC=90

∘

−θ

Now, ∠AED=90

∘

⇒∠AEC+∠DEC=90

∘

⇒∠DEC=90

∘

−(90

∘

−θ)=θ

Thus, in Δ ACE and DCE, we have

∠CAE=∠CED=θand∠ACE=∠ECD=90

∘

So, by AA similarly criterion, we have

ΔACE ΔECD

⇒

EC

AC

=

CD

CE

⇒

CF−EF

AC

=

BC−BD

CF−EF

⇒

R−5

R

=

R−9

R−5

⇒R(R−9)=(R−5)

2

⇒0=−R+25⇒R=25cm

Substituting the value of R in (i), we get

25−r=4.5 ⇒ r=20.5cm

Answer 2

Answer:

$\huge\underline\red{Answer}$

Let the radii of the larger and smaller circles be R and r respectively. Then,

BD=9cm

⇒ 2R−2r=9

⇒ R−r=4.5

Join AE and DE. Let ∠CAE=θ Then, ∠AEC=90∘−θ

Now, ∠AED=90°

⇒∠AEC+∠DEC=90 ∘

⇒∠DEC=90 ∘ −(90 ∘−θ)=θ

Now, ∠AED=90 ∘

⇒∠AEC+∠DEC=90∘

⇒∠DEC=90∘ −(90∘−θ)=θ

Thus, in Δ ACE and DCE, we have

∠CAE=∠CED=θand∠ACE=∠ECD=90 ∘

So, by AA similarly criterion, we have

ΔACE ΔECD

⇒ AC/EC = CE/CD

⇒ AC/CF - EF = CF - EF/BC - BDB

⇒ R/R - 5 = R - 5/R - 9

R(R - 9) = (R - 5)²

R² - 9R = R² + (-5)² - 2 × R × -5 [R² gets cancelled]

-9R = 25 + 10R

-19R = 25

R = -25/19

Substituting the value of R in (i), we get

= -25/19 - r

= -25/19 - 4/1

= -25 - 76/19

= -51/19