Question-7544

Question Number 7544 by Tawakalitu. last updated on 03/Sep/16

Commented by sou1618 last updated on 03/Sep/16

x = O A

O B = \sqrt{x^{2} + 144}

Δ A B O = \frac{1}{2} 12 x

Δ A O B = \frac{1}{2} 3 (12 + x + \sqrt{x^{2} + 144})

s o

12 x = 3 (12 + x + \sqrt{x^{2} + 144})

\sqrt{x^{2} + 144} = 3 x - 12

x^{2} + 144 = 9 x^{2} - 72 x + 144

x (72 - 8 x) = 0

x = 9

A B : A O = C E : C B

12 : 9 = C E : 12

C E = 16

B E = \sqrt{16^{2} + 12^{2}} = 4 \times 5 = 20

Δ B C E = \frac{1}{2} 12 \times 16 = 96

Δ B C E = \frac{1}{2} r (12 + 16 + 20) = 24 r

s o

96 = 24 r

r = 4

Commented by sou1618 last updated on 03/Sep/16

Commented by Tawakalitu. last updated on 03/Sep/16

W o w, i r e a l l y a p p r e c i a t e s i r . G o d b l e s s y o u .

Commented by Rasheed Soomro last updated on 03/Sep/16

I {d i d n}^{'} t u n d e r s t a n d h o w Δ A O B = \frac{1}{2} 3 (12 + x + \sqrt{x^{2} + 144}) ?

Commented by sandy_suhendra last updated on 03/Sep/16

t h e s m a l l e r c i r c l e i s c a l l e d t h e i n c i r c l e o f △ A O B

t h e r a d i u s o f a n i n c i r c l e : r = \frac{A}{s} \Rightarrow A = r s

w h e r e A i s t h e a r e a △ A O B a n d s i s s e m i p e r i m e t e r o f △ A O B

s o s = \frac{1}{2} (A O + A B + B O)

Commented by Rasheed Soomro last updated on 03/Sep/16

TH α n k S!

Commented by sandy_suhendra last updated on 03/Sep/16

u s i n g y o u r p i c t u r e a n d y o u r a n s w e r f o r A O = 9

a n d t h e r u l e o f s i m i l a r i t y \Rightarrow △ A O B \sim △ B C E

s o r : A O = R : B C \Rightarrow r = r a d i u s o f s m a l l e r c i r c l e

R = r a d i u s o f b i g g e r c i r c l e

3 : 9 = R : 12 \Rightarrow R = 4