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Question Number 190564 by Best1 last updated on 06/Apr/23

	Answered by a.lgnaoui last updated on 06/Apr/23

	$∙ 1 a p e r i m e t e r e (p)$ $[(2, 8 \times 2) + 1, 6] + \frac{1, 6 \times π}{2}$ $= 7, 1 + \frac{4 π}{5} \Rightarrow p = 9, 61 m$ $∙ 2 a A r e a$ $= (2, 8 \times 1, 6) + π {(0, 8)}^{2}$ $A = 6, 49 m^{2}$ $1 b p e r i m e t r e$ $p = (68 \times 2) + 25 π c m$ $2 b A r e a = 68 \times 50 + π \times {(25)}^{2}$ $= 5 363, 5 {c m}^{2}$ $= 5, 364 m^{2}$ $(t o f o l l o w f o r t h e r e s t . . .)$
	Answered by a.lgnaoui last updated on 06/Apr/23

	$e) l e t A_{1} = A r e a C_{1} [R_{1} = 15 c m]$ $A_{2} : A r e a C_{2} [R_{2} = 5 c m]$ $A_{3} : A r e a C 3 [R_{3} = R_{2}]$ $A_{4} : A r e a C_{4} [R_{4} = R_{2}]$ $A = \frac{π (R_{1}^{2} - R_{2}^{2})}{2} + π R_{2}^{2}$ $= \frac{π R_{1}^{2}}{2} + \frac{π R_{2}^{2}}{2} = \frac{π}{2} (R_{1}^{2} + R_{2}^{2})$ $= \frac{π}{2} (15^{2} + 5^{2})$ $A r e a = 392, 7$
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