Question-103171

Question Number 103171 by bemath last updated on 13/Jul/20

Commented by bobhans last updated on 13/Jul/20

(2 a) a s e c t o r o f a c i r c l e w i t h r a d i u s r h a s

p e r i m e t e r 20 u n i t s a n d a r e a 21 s q u a r e

u n i t s . F i n d a l l p o s s i b l e o f r a n d t h e

c o r r e s p o n d i n g v a l u e o f θ .

Answered by bobhans last updated on 13/Jul/20

p e r i m e t e r = 2 r + θ r

20 = 2 r + θ r \Rightarrow θ = \frac{20 - 2 r}{r} \dots (1)

a r e a = \frac{1}{2} θ r^{2}

21 = \frac{1}{2} θ r^{2} \dots (2)

(1) a n d (2) \Rightarrow 42 = (\frac{20 - 2 r}{r}) . r^{2}

42 = 20 r - 2 r^{2} \Rightarrow r^{2} - 10 r + 21 = 0

(r - 3) (r - 7) = 0 {\begin{cases} r = 3 u n i t \\ r = 7 u n i t \end{cases}

f o r r = 3 u n i t \Rightarrow θ = \frac{20 - 6}{3} = \frac{14}{3} r a d i a n

f o r r = 7 u n i t \Rightarrow θ = \frac{20 - 14}{7} = \frac{6}{7} r a d i a n

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