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Question Number 125084 by ajfour last updated on 08/Dec/20

Commented by ajfour last updated on 08/Dec/20

Find x, hence radius of circle.

Findx,henceradiusofcircle.

Commented by ajfour last updated on 08/Dec/20

Commented by mr W last updated on 08/Dec/20

x can be any value, because the shape formed by both rectangles is always cyclic.

xcanbeanyvalue,becausetheshapeformedbybothrectanglesisalwayscyclic.

Answered by mr W last updated on 08/Dec/20

Commented by mr W last updated on 08/Dec/20

ABCD is always cyclic. AB=CD=(√((((x−1)/2))^2 +(1+x)^2 )) AC=BD=(√((((x+1)/2))^2 +(1+x)^2 ))=(((x+1)(√5))/2) R=radius Δ_(ABD) =((1×(1+x))/2) formula R=((abc)/(4Δ)) ⇒R=((√([(((x−1)/2))^2 +(1+x)^2 ][(((x+1)/2))^2 +(1+x)^2 ]))/(2(1+x))) ⇒R=((√(5(5x^2 +6x+5)))/8)

ABCDisalwayscyclic.AB=CD=(x−12)2+(1+x)2AC=BD=(x+12)2+(1+x)2=(x+1)52R=radiusΔABD=1×(1+x)2formulaR=abc4Δ⇒R=[(x−12)2+(1+x)2][(x+12)2+(1+x)2]2(1+x)⇒R=5(5x2+6x+5)8

Commented by ajfour last updated on 08/Dec/20

Very Nice solution, Sir!

VeryNicesolution,Sir!

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