Question-224773

Question Number 224773 by Ismoiljon_008 last updated on 03/Oct/25

Answered by mr W last updated on 03/Oct/25

Commented by mr W last updated on 03/Oct/25

CD=(√(23^2 −9^2 ))=8(√7) PB=2R−5 DQ=2R CQ=(√((2R)^2 −(8(√7))^2 ))=2(√(R^2 −112)) PQ=2(√(R^2 −112))−9 9×(2(√(R^2 −112))−9)=5×(2R−5) 9(√(R^2 −112))=5R+28 81(R^2 −112)=25R^2 +280R+28^2 R^2 −5R−176=0 (R−16)(R+11)=0 ⇒R=16 ✓

$${CD}=\sqrt{\mathrm{23}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} }=\mathrm{8}\sqrt{\mathrm{7}} \\ $$$${PB}=\mathrm{2}{R}−\mathrm{5} \\ $$$${DQ}=\mathrm{2}{R} \\ $$$${CQ}=\sqrt{\left(\mathrm{2}{R}\right)^{\mathrm{2}} −\left(\mathrm{8}\sqrt{\mathrm{7}}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{{R}^{\mathrm{2}} −\mathrm{112}} \\ $$$${PQ}=\mathrm{2}\sqrt{{R}^{\mathrm{2}} −\mathrm{112}}−\mathrm{9} \\ $$$$\mathrm{9}×\left(\mathrm{2}\sqrt{{R}^{\mathrm{2}} −\mathrm{112}}−\mathrm{9}\right)=\mathrm{5}×\left(\mathrm{2}{R}−\mathrm{5}\right) \\ $$$$\mathrm{9}\sqrt{{R}^{\mathrm{2}} −\mathrm{112}}=\mathrm{5}{R}+\mathrm{28} \\ $$$$\mathrm{81}\left({R}^{\mathrm{2}} −\mathrm{112}\right)=\mathrm{25}{R}^{\mathrm{2}} +\mathrm{280}{R}+\mathrm{28}^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} −\mathrm{5}{R}−\mathrm{176}=\mathrm{0} \\ $$$$\left({R}−\mathrm{16}\right)\left({R}+\mathrm{11}\right)=\mathrm{0} \\ $$$$\Rightarrow{R}=\mathrm{16}\:\:\checkmark \\ $$

Commented by JV2BTC last updated on 03/Oct/25

$$ \\ $$$$\mathrm{ensured}\:\mathrm{by}\:\mathrm{the}\:\mathrm{Thaless}\:\mathrm{theorem}\: \\ $$

Commented by Ismoiljon_008 last updated on 04/Oct/25

$$\:{Thank}\:{you} \\ $$

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