A-chord-divides-the-circle-in-two-segments-having-areas-s-1-and-s-2-If-diameter-perpendicular-to-this-chord-is-cut-into-1-3-by-the-chord-what-is-s-1-s-2-

Question Number 3808 by Rasheed Soomro last updated on 21/Dec/15

A c h o r d d i v i d e s t h e c i r c l e i n t w o

s e g m e n t s, h a v i n g a r e a s s_{1} a n d s_{2} .

I f d i a m e t e r, p e r p e n d i c u l a r t o t h i s

c h o r d i s c u t i n t o 1 : 3 b y t h e c h o r d, w h a t i s s_{1} : s_{2} ?

Commented by Rasheed Soomro last updated on 21/Dec/15

L e t A B i s a c h o r d a n d C D (⊥ A B) i s a d i a m e t e r .

L e t A B c u t s C D a t p o i n t O

I f C O : O D = 1 : 3, w h a t i s s_{1} : s_{2}

s_{1} a n d s_{2} a r e a r e a s o f t w o s e g m e n t s o f t h e

c i r c l e p r o d u c e d b y c h o r d A B .

Answered by Yozzii last updated on 21/Dec/15

(I n t e g r a l C a l c u l u s c a n a l t e r n a t i v e l y

b e u s e d t o f i n d t h e v a l u e o f \frac{s_{1}}{s_{2}} .)

D r a w a c i r c l e w i t h d i a m e t e r l e n g t h d a n d

c e n t r e Q . A d i a m e t e r o f t h e c i r c l e n o r m a l l y

c u t s a c h o r d A B o n t h e c i r c l e s u c h t h a t

d i s p a r t i t i o n e d i n t h e r a t i o 1 : 3 . L e t

L F b e s u c h a d i a m e t e r a n d D b e t h e

p o i n t o f i n t e r s e c t i o n o f t h e c h o r d A B

a n d d i a m e t e r L F . W e t h u s h a v e

∣ L D ∣= 3 a a n d ∣ D F ∣= a (a > 0) s o t h a t D F : L D = 1 : 3 .

T h e r e f o r e, d = 3 a + a = 4 a \Rightarrow r a d i u s r = 2 a .

C o m p l e t e t h e i s o s c e l e s t r i a n g l e △ A Q B

a n d l e t ∠ A Q B = θ . T h e a r e a A_{1} o f t h e s e c t o r A B

o f t h e c i r c l e i s t h e n

A_{1} = \frac{1}{2} r^{2} θ = \frac{1}{2} \times 4 a^{2} θ = 2 a^{2} θ .

T h e a r e a o f △ A Q B, d e n o t e d b y A_{2}, i s g i v e n b y

A_{2} = \frac{1}{2} 2 a \times 2 a \times s i n θ = 2 a^{2} s i n θ .

N o w, l e t s_{1} b e t h e a r e a b e t w e e n t h e c h o r d

A B a n d t h e a r c A B .

\Rightarrow s_{1} = A_{1} - A_{2} = 2 a^{2} (θ - s i n θ)

H e n c e, t h e a r e a o f t h e o t h e r s e g m e n t

o f t h e c i r c l e i s s_{2} = π r^{2} - s_{1} .

s_{2} = 4 π a^{2} - 2 a^{2} (θ - s i n θ)

s_{2} = 2 a^{2} (2 π + s i n θ - θ)

T h e r e f o r e, \frac{s_{1}}{s_{2}} = \frac{θ - s i n θ}{2 π + s i n θ - θ} .

R e t u r n i n g t o △ A Q B, w e c a n y i e l d

a r i g h t - a n g l e d t r i a n g l e △ A Q D

w i t h ∠ A Q D = \frac{1}{2} ∠ A Q B = \frac{θ}{2}

a n d ∠ Q D A = \frac{π}{2} .

W e t h e n h a v e c o s \frac{θ}{2} = \frac{Q D}{A Q} = \frac{a}{2 a} = \frac{1}{2}

\Rightarrow \frac{θ}{2} = \frac{π}{3} \Rightarrow θ = \frac{2 π}{3} .

∴ \frac{s_{1}}{s_{2}} = \frac{\frac{2 π}{3} - s i n \frac{2 π}{3}}{2 π + s i n \frac{2 π}{3} - \frac{2 π}{3}} = \frac{\frac{2 π}{3} - \frac{\sqrt{3}}{2}}{2 π + \frac{\sqrt{3}}{2} - \frac{2 π}{3}}

\frac{s_{1}}{s_{2}} = \frac{4 π - 3 \sqrt{3}}{8 π + 3 \sqrt{3}}

Commented by Rasheed Soomro last updated on 21/Dec/15

^{″} \dots . A d i a m e t e r o f t h e c i r c l e n o r m a l l y

c u t s a c h o r d A B \underline{o n t h e c i r c l e} s u c h t h a t

d i s p a r t i t i o n e d i n t h e r a t i o 1 : 3 \dots .^{″}

W h a t d o y o u m e a n b y t h i s ?

Commented by Yozzii last updated on 21/Dec/15

n o r m a l l y \Rightarrow p e r p e n d i c u l a r l y (i n m a t h t e r m s)

Commented by Yozzii last updated on 21/Dec/15

T h e d i a m e t e r c u t s t h e c h o r d a t 90 °

w i t h t h e c h o r d d i v i n d i n g t h e l e n g t h o f

t h e d i a m e t e r i n t h e r a t i o 1 : 3 .

Commented by Yozzii last updated on 21/Dec/15

T h e q u e s t i o n {h a d n}^{'} t s p e c i f i e d w h e t h e r

s_{1} > s_{2} o r s_{2} > s_{1} s o t r u l y \frac{s_{1}}{s_{2}} c a n t a k e

t w o v a l u e s t h a t a r e r e c i p r o c a l s

o f e a c h o t h e r .

Commented by Rasheed Soomro last updated on 22/Dec/15

T h a n k s! M i s u n d e r s t a n d i n g i n u n d e r s t a n d i n g

l a n g u a g e . I g o t^{'} \boldsymbol {n o r m a l l y}^{'} i n n o r m a l l a n g u a g e,

n o t a s m a t h e m a t i c a l t e r m .

Commented by Rasheed Soomro last updated on 21/Dec/15

G^{O^{V} O} D A p p r o a c h!

Answered by Yozzii last updated on 21/Dec/15

(I n t e g r a l c a l c u l u s m e t h o d b a s e d o n

i n i t i a l i n f o r m a t i o n f r o m f i r s t a n s w e r .)

D e f i n e a c i r c l e C w i t h e q u a t i o n

x^{2} + y^{2} = 4 a^{2}, a > 0 .

F o r C, l e t s_{1} b e t h e a r e a b e t w e e n t h e

c h o r d A B a n d t h e a r c A B . W e c a n l e t

t h e d i a m e t e r n o r m a l t o A B b e c o n c u r r e n t

w i t h t h e x - a x i s .

∴ s_{1} = \int_{a}^{2 a} y d x

s_{1} = \int_{a}^{2 a} \sqrt{4 a^{2} - x^{2}} d x

L e t x = 2 a s i n θ \Rightarrow d x = 2 a c o s θ d θ

a n d \sqrt{4 a^{2} - x^{2}} = 2 a c o s θ .

A t x = a \Rightarrow s i n θ = 1 / 2 \Rightarrow θ = \frac{π}{6} .

A t x = 2 a \Rightarrow s i n θ = 1 \Rightarrow θ = π / 2 .

∴ s_{1} = 2 \int_{π / 6}^{π / 2} 4 a^{2} {c o s}^{2} θ d θ

s_{1} = 4 a^{2} \int_{π / 6}^{π / 2} (1 + c o s 2 θ) d θ

s_{1} = 4 a^{2} (θ + \frac{1}{2} s i n 2 θ) ∣_{π / 6}^{π / 2}

s_{1} = 4 a^{2} (\frac{π}{2} - \frac{π}{6} + \frac{1}{2} s i n π - \frac{1}{2} s i n \frac{π}{3})

s_{1} = 4 a^{2} (\frac{π}{3} - \frac{\sqrt{3}}{4})

s_{1} = \frac{a^{2}}{3} (4 π - 3 \sqrt{3})

T h e a r e a s_{2} o f t h e o t h e r s e g m e n t i s

g i v e n b y s_{2} = π r^{2} - s_{1} .

∴ s_{2} = 4 π a^{2} - 4 a^{2} \frac{4 π - 3 \sqrt{3}}{12}

s_{2} = 4 a^{2} (π - \frac{4 π - 3 \sqrt{3}}{12})

s_{2} = 4 a^{2} (\frac{12 π - 4 π + 3 \sqrt{3}}{12})

s_{2} = \frac{a^{2}}{3} (8 π + 3 \sqrt{3})

∴ \frac{s_{1}}{s_{2}} = \frac{4 π - 3 \sqrt{3}}{8 π + 3 \sqrt{3}}

Commented by Rasheed Soomro last updated on 21/Dec/15

A n O \boldsymbol ther

\overset{O^{V} O}{G D}

Approach!