Question-167633

Question Number 167633 by cherokeesay last updated on 21/Mar/22

Commented by som(math1967) last updated on 22/Mar/22

Commented by som(math1967) last updated on 22/Mar/22

B L t a n j e n t O L r a d i u s ∴ B L ⊥ O L

∴ ∠ B L O = 90 ∴ ∠ B O L = 90 - 60 = 30

[∵ △ A B C e q u i l a t e r a l \Rightarrow ∠ L B O = 60

l e t O L = r

∴ L M = r s i n 30 = \frac{r}{2}

M O = r c o s 30 = \frac{\sqrt{3} r}{2}

l e n g t h o f r e c t a n g l e = 2 \times M O = \sqrt{3} r

a r e a o f r e c t a n g l e = \frac{\sqrt{3} r^{2}}{2} s q u n i t

a r e a o f h a l f d i s c = \frac{1}{2} \times π \times r^{2}

∴ \frac{r e c t a g l e a r e a}{h a l f d i s c a r e a}

= \frac{\frac{\sqrt{3} r^{2}}{2}}{\frac{π r^{2}}{2}} = \frac{\sqrt{3}}{π}

Commented by som(math1967) last updated on 21/Mar/22

\frac{r e c t a n g l e a r e a}{h a l f d i s c a r e a} = \frac{\sqrt{3}}{π} ?

Commented by cherokeesay last updated on 21/Mar/22

y e s S i r!

t h a n k s .

Commented by Lakers last updated on 21/Mar/22

y o u r w o r k i n g p l e a s e

Commented by Tawa11 last updated on 22/Mar/22

Nice sir

Answered by cherokeesay last updated on 22/Mar/22

\frac{R}{C H} = c o s 30 ° = \frac{\sqrt{3}}{2} \Leftrightarrow R = \frac{C H \sqrt{3}}{2}

\frac{a}{R} = c o s 30 ° = \frac{\sqrt{3}}{2} \Leftrightarrow a = \frac{3 C H}{4}

\frac{b}{R} = s i n 30 ° = \frac{1}{2} \Leftrightarrow b = \frac{C H \sqrt{3}}{4}

A_{r e c .} = 2 a b = \frac{6 {C H}^{2} \sqrt{3}}{16}

\frac{A_{r e c .}}{A_{h . d}} = \frac{\frac{6 {C H}^{2} \sqrt{3}}{16}}{\frac{π {(\frac{C H \sqrt{3}}{2})}^{2}}{2}} = \frac{6 {C H}^{2} \sqrt{3}}{16} \times \frac{8}{3 π {C H}^{2}} =

\frac{\sqrt{3}}{π}