Observe that OE, OF , OG and OH perpendicular to AB, BC, CD and AD. Draw diameter FF’ We have ∠ (EOF’)= ∠ (B) … ( both angles supplement to ∠ (EOF) ) ∠ (HOG)= ∠ (B)…. ( both angles supplement to∠ (D)) Triangles EOF’ congruent to HOG ( case SAS) => EF’= HG In right triangle FEF’ we have EF^2+EF’^2= EF^2+HG^2= 36+16= 52= 4. Radius^2 Similarly FG^2+EH^2= 4. Radius^2= 25+x^2 So x^2= 52-25=27 and x=3.sqrt(3)
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ReplyDeleteObserve that OE, OF , OG and OH perpendicular to AB, BC, CD and AD.
Draw diameter FF’
We have ∠ (EOF’)= ∠ (B) … ( both angles supplement to ∠ (EOF) )
∠ (HOG)= ∠ (B)…. ( both angles supplement to∠ (D))
Triangles EOF’ congruent to HOG ( case SAS) => EF’= HG
In right triangle FEF’ we have EF^2+EF’^2= EF^2+HG^2= 36+16= 52= 4. Radius^2
Similarly FG^2+EH^2= 4. Radius^2= 25+x^2
So x^2= 52-25=27 and x=3.sqrt(3)
Notice that GE_|_FH, thus EF^2+GH^2=FG^2+EH^2, wherefrom x^2=27.
ReplyDeleteBest regards