A-semicircle-contains-a-square-of-possible-largest-area-If-s-is-the-measure-of-the-side-of-the-square-what-is-the-radius-of-the-semicircle-

Question Number 3840 by Rasheed Soomro last updated on 22/Dec/15

A s e m i c i r c l e c o n t a i n s a s q u a r e o f

p o s s i b l e l a r g e s t a r e a . I f s i s t h e m e a s u r e

o f t h e s i d e o f t h e s q u a r e, w h a t i s t h e

r a d i u s o f t h e s e m i c i r c l e ?

Answered by Yozzii last updated on 24/Dec/15

D r a w a s e m i c i r c l e o f r a d i u s r . I f t h e

s e m i c i r c l e c o n t a i n s a s q u a r e o f

m a x i m u m a r e a, t h e c e n t r e o f t h e s t r a i g h t e d g e

o f t h e s e m i c i r c l e l i e s a t t h e m i d - p o i n t o f

t h e s i d e o f t h e s q u a r e c o n c u r r e n t w i t h

t h e s t r a i g h t e d g e o f t h e s e m i c i r c l e .

A l s o, t h e r e m a i n i n g t w o c o r n e r s o f t h e

s q u a r e t o u c h t h e a r c o f t h e s e m i c i r c l e .

C a l l t h e m i d p o i n t p o i n t O a n d l a b e l t h e s q u a r e

A B C D . C D i s t h e s i d e w i t h O o n i t

a n d A B i s o p p o s i t e C D . F o r m t h e

t r i a n g l e △ O B C . B y v i r t u e o f t h e

s y m m e t r y o f t h e s q u a r e, a r e a o f t h e

s q u a r e A_{1} = 4 \times {a r e a o f △ O B C} . T h e

s q u a r e h a s s i d e l e n g t h s

∴ s^{2} = 4 {a r e a o f △ O B C} .

L e t θ = ∠ B O C . \Rightarrow O C = r c o s θ .

∴ a r e a o f △ O B C = \frac{1}{2} O C \times B C = \frac{s r c o s}{2} .

∴ s^{2} = 2 s r c o s θ \Rightarrow c o s θ = \frac{s}{2 r} .

F r o m △ O B C, s i n θ = \frac{s}{r} .

∵ {s i n}^{2} θ + {c o s}^{2} θ = 1

\Rightarrow \frac{s^{2}}{r^{2}} + \frac{s^{2}}{4 r^{2}} = 1 \Rightarrow 4 s^{2} + s^{2} = 4 r^{2}

r^{2} = \frac{5}{4} s^{2} \Rightarrow r = \frac{\sqrt{5}}{2} s (r, s > 0)

Commented by Rasheed Soomro last updated on 24/Dec/15

G_{O}^{O} D_{E x p l a n a t i o n}^{L o g i c}!