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Question Number 191715 by Shlock last updated on 29/Apr/23

Answered by mr W last updated on 29/Apr/23

Commented by Shlock last updated on 29/Apr/23

Sir,canyon use the similarity of the right triangles? Can you give more explanation in details?

Commented by mr W last updated on 29/Apr/23

R=radius of semi−circle (√((R−a)^2 −a^2 ))+(√((R−b)^2 −b^2 ))=a+b (√(R(R−2a)))=a+b−(√(R(R−2b))) R(R−2a)=(a+b)^2 +R(R−2b)−2(a+b)(√(R(R−2b))) (a+b)^2 −2(b−a)R=2(a+b)(√(R(R−2b))) (a+b)^2 −4(b−a)(a+b)^2 R+4(b−a)^2 R^2 =4(a+b)^2 R^2 −8b(a+b)^2 R 16abR^2 −4(a+b)^3 R−(a+b)^4 =0 R=(((a+b)^2 [a+b+(√((a+b)^2 +4ab))])/(8ab)) AB=2R ⇒AB=(((a+b)^2 [a+b+(√((a+b)^2 +4ab))])/(4ab)) with a=4, b=6 AB=((100(10+(√(100+4×24))))/(4×24))=25

R=radiusofsemi−circle(R−a)2−a2+(R−b)2−b2=a+bR(R−2a)=a+b−R(R−2b)R(R−2a)=(a+b)2+R(R−2b)−2(a+b)R(R−2b)(a+b)2−2(b−a)R=2(a+b)R(R−2b)(a+b)2−4(b−a)(a+b)2R+4(b−a)2R2=4(a+b)2R2−8b(a+b)2R16abR2−4(a+b)3R−(a+b)4=0R=(a+b)2[a+b+(a+b)2+4ab]8abAB=2R⇒AB=(a+b)2[a+b+(a+b)2+4ab]4abwitha=4,b=6AB=100(10+100+4×24)4×24=25

Commented by Shlock last updated on 29/Apr/23

Perfect ��

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