Math Question and Answers


Question and Answers Forum

All Questions Topic List
Mensuration Questions
Previous in All Question Next in All Question
Previous in Mensuration Next in Mensuration
Question Number 187932 by Rupesh123 last updated on 23/Feb/23

	Answered by a.lgnaoui last updated on 24/Feb/23
	$O_1 :centre du cercle(C_1 )= origine Ref Cercles( C_1 )et(C_2 )d equations { ((x^2 +y_1 ^2 =100 (1))),(((x−10)^2 +y_2 ^2 =49 (2))) :} Ponts communs(M; N )definies par y_1 =y_2 ⇒ x=((151)/(20)) O_1 H=7,55 y=(√(100−((151^2 )/(20^2 )))) =±(√(43)) MH=O_1 Q=6,55724 Aire(Bleu) limite par C_1 erC_2 A=A_1 +A_2 tan α=((OQ)/(OH))= ((20(√(43)))/(151))⇒ [α=41°] Aire[A_1 ]=[αR_1 ^2 −O_1 H ×MH] α=((41π)/(180)) A_1 =((5×41π)/9)−49,5) Aire A_1 =22 Aire A_2 =ϕR_2 ^2 −O_2 H×MH O_2 H=OO−O_1 H=2,45⇒ tan ϕ=((MH)/(O_2 H)) ⇒ϕ≅ ((23π)/(60)) ⇒ AireA_2 =((49×23π)/(60)) −16=43 A=A_1 +A_2 Aire Totale=65 Totale generale Aire= 65+(αR_1 ^2 +ϕR_2 ^2 ) =65+131=196$
	$O_{1} : c e n t r e d u c e r c l e (C_{1}) = o r i g i n e R e f$ $C e r c l e s (C_{1}) e t (C_{2}) d e q u a t i o n s$ ${\begin{cases} x^{2} + y_{1}^{2} = 100 (1) \\ {(x - 10)}^{2} + y_{2}^{2} = 49 (2) \end{cases}$ $P o n t s c o m m u n s (M; N) d e f i n i e s p a r$ $y_{1} = y_{2} \Rightarrow x = \frac{151}{20} O_{1} H = 7, 55$ $y = \sqrt{100 - \frac{151^{2}}{20^{2}}} = \pm \sqrt{43} M H = O_{1} Q = 6, 55724$ $A i r e (B l e u) l i m i t e p a r C_{1} {e r C}_{2}$ $A = A_{1} + A_{2}$ $\tan α = \frac{O Q}{O H} = \frac{20 \sqrt{43}}{151} \Rightarrow [α = 41 °]$ $A i r e [A_{1}] = [α R_{1}^{2} - O_{1} H \times M H] α = \frac{41 π}{180}$ $A_{1} = \frac{5 \times 41 π}{9} - 49, 5) A i r e A_{1} = 22$ $A i r e A_{2} = φ R_{2}^{2} - O_{2} H \times M H$ $O_{2} H = O O - O_{1} H = 2, 45 \Rightarrow \tan φ = \frac{M H}{O_{2} H}$ $\Rightarrow φ ≅ \frac{23 π}{60} \Rightarrow A {i r e A}_{2} = \frac{49 \times 23 π}{60} - 16 = 43$ $A = A_{1} + A_{2}$ $A i r e T o t a l e = 65$ $T o t a l e g e n e r a l e$ $A i r e = 65 + (α R_{1}^{2} + φ R_{2}^{2})$ $= 65 + 131 = 196$
	Commented by a.lgnaoui last updated on 25/Feb/23

	Answered by mr W last updated on 24/Feb/23

	Commented by mr W last updated on 24/Feb/23

	$\sin γ = \cos β = \frac{7}{20}$ $\sin 2 β = 2 \times \frac{7}{20} \times \frac{\sqrt{20^{2} - 7^{2}}}{20} = \frac{21 \sqrt{39}}{200}$ $α = 2 γ$ $\cos α = 1 - 2 \times {(\frac{7}{20})}^{2} = \frac{151}{200}$ $\sin 2 α = 2 \times \frac{151}{200} \times \frac{\sqrt{200^{2} - 151^{2}}}{200} = \frac{3171 \sqrt{39}}{20000}$ $A_{b l u e} = \frac{10^{2}}{2} (2 α - \sin 2 α) + \frac{7^{2}}{2} (2 β - \sin 2 β)$ $= \frac{10^{2}}{2} (4 \sin^{- 1} \frac{7}{20} - \frac{3171 \sqrt{39}}{20000}) + \frac{7^{2}}{2} (2 \cos^{- 1} \frac{7}{20} - \frac{21 \sqrt{39}}{200})$ $\approx 65.389 778$
	Commented by Rupesh123 last updated on 01/Mar/23
	Excellent!
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum