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A-quadrilateral-with-consecutive-sides-of-lenght-7-15-15-and-d-is-inscribed-in-a-circle-with-its-diameter-d-Find-the-radius-of-circle-




Question Number 11759 by Joel576 last updated on 31/Mar/17
A quadrilateral with consecutive sides of lenght 7, 15, 15, and d  is inscribed in a circle with its diameter d.  Find the radius of circle
$$\mathrm{A}\:\mathrm{quadrilateral}\:\mathrm{with}\:\mathrm{consecutive}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{lenght}\:\mathrm{7},\:\mathrm{15},\:\mathrm{15},\:\mathrm{and}\:{d} \\ $$$$\mathrm{is}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{its}\:\mathrm{diameter}\:{d}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{circle} \\ $$
Answered by ajfour last updated on 31/Mar/17
Commented by ajfour last updated on 31/Mar/17
AB=7 ;    BC=15;   CD=15;  AD= d =2r  OM= (7/2)   midpoint theorem.  CM= (r−(7/2))  ;    DM= ((√(4r^2  −49))/2)  CM^2  + DM^2  = CD^2   (r−(7/2))^2 + (((4r^2  −49)/4)) = 225  2r^2 −7r −225 =0  r =((7±(√(49+1800)))/4)  and as r>0 ,  r= ((7+43)/4) = ((25)/2) .
$${AB}=\mathrm{7}\:;\:\:\:\:{BC}=\mathrm{15};\:\:\:{CD}=\mathrm{15}; \\ $$$${AD}=\:{d}\:=\mathrm{2}{r} \\ $$$${OM}=\:\frac{\mathrm{7}}{\mathrm{2}}\:\:\:{midpoint}\:{theorem}. \\ $$$${CM}=\:\left({r}−\frac{\mathrm{7}}{\mathrm{2}}\right)\:\:;\:\:\:\:{DM}=\:\frac{\sqrt{\mathrm{4}{r}^{\mathrm{2}} \:−\mathrm{49}}}{\mathrm{2}} \\ $$$${CM}^{\mathrm{2}} \:+\:{DM}^{\mathrm{2}} \:=\:{CD}^{\mathrm{2}} \\ $$$$\left({r}−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} +\:\left(\frac{\mathrm{4}{r}^{\mathrm{2}} \:−\mathrm{49}}{\mathrm{4}}\right)\:=\:\mathrm{225} \\ $$$$\mathrm{2}{r}^{\mathrm{2}} −\mathrm{7}{r}\:−\mathrm{225}\:=\mathrm{0} \\ $$$${r}\:=\frac{\mathrm{7}\pm\sqrt{\mathrm{49}+\mathrm{1800}}}{\mathrm{4}}\:\:{and}\:{as}\:{r}>\mathrm{0}\:, \\ $$$${r}=\:\frac{\mathrm{7}+\mathrm{43}}{\mathrm{4}}\:=\:\frac{\mathrm{25}}{\mathrm{2}}\:. \\ $$$$ \\ $$
Commented by Joel576 last updated on 31/Mar/17
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$
Commented by Joel576 last updated on 01/Apr/17
but how did you know that point C, M, O  in one line ?
$$\mathrm{but}\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{know}\:\mathrm{that}\:\mathrm{point}\:{C},\:{M},\:{O} \\ $$$$\mathrm{in}\:\mathrm{one}\:\mathrm{line}\:? \\ $$
Commented by ajfour last updated on 01/Apr/17
△CBD is isosceles so perpendicular  from C on BD bisects it.A line   perpendicular to BD from center  of circle even bisects its chord BD.
$$\bigtriangleup{CBD}\:{is}\:{isosceles}\:{so}\:{perpendicular} \\ $$$${from}\:{C}\:{on}\:{BD}\:{bisects}\:{it}.{A}\:{line}\: \\ $$$${perpendicular}\:{to}\:{BD}\:{from}\:{center} \\ $$$${of}\:{circle}\:{even}\:{bisects}\:{its}\:{chord}\:{BD}. \\ $$
Commented by mrW1 last updated on 01/Apr/17
what′s the result if the sides are 15,7,15 and d?
$${what}'{s}\:{the}\:{result}\:{if}\:{the}\:{sides}\:{are}\:\mathrm{15},\mathrm{7},\mathrm{15}\:{and}\:{d}? \\ $$
Commented by ajfour last updated on 01/Apr/17
the same.
$${the}\:{same}. \\ $$

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