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Question Number 201907 by mnjuly1970 last updated on 15/Dec/23

	Answered by Rasheed.Sindhi last updated on 15/Dec/23

	$∙ R a d i u s o f S e m i - c i r c l e R = \frac{10}{2} = 5 c m$ $∙ l e t r a d o u s o f g r e a t e r q u a r t e r - c i r c l e = r_{1}$ ${(R + r_{1})}^{2} = 10^{2} + 5^{2} = 125$ $5 + r_{1} = 5 \sqrt{5}$ $r_{1} = 5 \sqrt{5} - 5 = 5 (\sqrt{5} - 1)$ $∙ l e t r_{2} i s r a d i u s o f s m a l l e r q u a r t e r c i r c l e$ $r_{2} = 10 - 5 (\sqrt{5} - 1) = 15 - 5 \sqrt{5}$ $∙ S e m i - c i r c l e - a r e a = \frac{π {(5)}^{2}}{2} = \frac{25 π}{2}$ $∙ G r e a t e r - q u a r t e r - c i r c l e - a r e a = \frac{π {(5 (\sqrt{5} - 1))}^{2}}{4}$ $∙ S m a l l e r - q u a r t e r - c i r c l e - a r e a = \frac{π {(15 - 5 \sqrt{5})}^{2}}{4}$ $B l u e - a r e a = (S q u a r e - a r e a) - (S e m i - c i r c l e - a r e a + G r e a t e r - q u a r t e r - c i r c l e - a r e a + S m a l l e r - q u a r t e r - c i r c l e - a r e a)$ $= 10^{2} - (\frac{25 π}{2} + \frac{π {(5 (\sqrt{5} - 1))}^{2}}{4} + \frac{π {(15 - 5 \sqrt{5})}^{2}}{4})$ $= 10^{2} - (\frac{50}{4} + \frac{{(5 (\sqrt{5} - 1))}^{2}}{4} + \frac{{(15 - 5 \sqrt{5})}^{2}}{4}) π$
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