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Question Number 72734 by Mr Jor last updated on 01/Nov/19
A triangle ABC is inscribed in a circle.AC=10cm,BC=7cm and AB=10cm.Find the radius of the circle.
$${A}\:{triangle}\:{ABC}\:{is}\:{inscribed}\:{in}\:{a} \\ $$$${circle}.{AC}=\mathrm{10}{cm},{BC}=\mathrm{7}{cm}\:{and}\: \\ $$$${AB}=\mathrm{10}{cm}.{Find}\:{the}\:{radius}\:{of}\:{the} \\ $$$${circle}. \\ $$
Answered by $@ty@m123 last updated on 01/Nov/19
a=7, b=10, c=10 s=((a+b+c)/2) =((27)/2) △=(√(s(s−a)(s−b)(s−c))) =(√(((27)/2)(((27)/2)−7)(((27)/2)−10)(((27)/2)−10))) =(√(((27)/2)×((13)/2)×(7/2)×(7/2))) =((21)/4)(√(39)) ......(1) R=((abc)/(4△)) =((7×10×10)/(4×((21(√(39)))/4))) =((100)/(3(√(39)))) Ans.
$${a}=\mathrm{7},\:{b}=\mathrm{10},\:{c}=\mathrm{10} \\ $$$${s}=\frac{{a}+{b}+{c}}{\mathrm{2}}\:=\frac{\mathrm{27}}{\mathrm{2}} \\ $$$$\bigtriangleup=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$$\:\:\:\:=\sqrt{\frac{\mathrm{27}}{\mathrm{2}}\left(\frac{\mathrm{27}}{\mathrm{2}}−\mathrm{7}\right)\left(\frac{\mathrm{27}}{\mathrm{2}}−\mathrm{10}\right)\left(\frac{\mathrm{27}}{\mathrm{2}}−\mathrm{10}\right)} \\ $$$$\:\:\:\:\:=\sqrt{\frac{\mathrm{27}}{\mathrm{2}}×\frac{\mathrm{13}}{\mathrm{2}}×\frac{\mathrm{7}}{\mathrm{2}}×\frac{\mathrm{7}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{21}}{\mathrm{4}}\sqrt{\mathrm{39}}\:\:\:……\left(\mathrm{1}\right) \\ $$$${R}=\frac{{abc}}{\mathrm{4}\bigtriangleup} \\ $$$$\:\:\:\:=\frac{\mathrm{7}×\mathrm{10}×\mathrm{10}}{\mathrm{4}×\frac{\mathrm{21}\sqrt{\mathrm{39}}}{\mathrm{4}}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{100}}{\mathrm{3}\sqrt{\mathrm{39}}}\:\:{Ans}. \\ $$
Answered by MJS last updated on 01/Nov/19
triangle with side lengths a, b, c circumcircle R=((abc)/δ) δ=(√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))
$$\mathrm{triangle}\:\mathrm{with}\:\mathrm{side}\:\mathrm{lengths}\:{a},\:{b},\:{c} \\ $$$$\mathrm{circumcircle}\:{R}=\frac{{abc}}{\delta} \\ $$$$\delta=\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)} \\ $$
Commented by $@ty@m123 last updated on 01/Nov/19
a+b+c=2s ∴δ=(√(2s.(2s−2a)(2s−2b)(2s−2c))) =4(√(s(s−a)(s−b)(s−c))) =4△
$${a}+{b}+{c}=\mathrm{2}{s} \\ $$$$\therefore\delta=\sqrt{\mathrm{2}{s}.\left(\mathrm{2}{s}−\mathrm{2}{a}\right)\left(\mathrm{2}{s}−\mathrm{2}{b}\right)\left(\mathrm{2}{s}−\mathrm{2}{c}\right)} \\ $$$$\:\:\:\:\:\:=\mathrm{4}\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$$\:\:\:\:\:=\mathrm{4}\bigtriangleup \\ $$
Answered by mr W last updated on 01/Nov/19
(√(10^2 −((7/2))^2 ))−R=(√(R^2 −((7/2))^2 )) 10^2 −2R(√(10^2 −((7/2))^2 ))=0 ⇒R=((10^2 )/( (√((2×10)^2 −7^2 ))))=((100)/( (√(27×13))))=((100(√(39)))/(117))≈5.338
$$\sqrt{\mathrm{10}^{\mathrm{2}} −\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} }−{R}=\sqrt{{R}^{\mathrm{2}} −\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{10}^{\mathrm{2}} −\mathrm{2}{R}\sqrt{\mathrm{10}^{\mathrm{2}} −\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow{R}=\frac{\mathrm{10}^{\mathrm{2}} }{\:\sqrt{\left(\mathrm{2}×\mathrm{10}\right)^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }}=\frac{\mathrm{100}}{\:\sqrt{\mathrm{27}×\mathrm{13}}}=\frac{\mathrm{100}\sqrt{\mathrm{39}}}{\mathrm{117}}\approx\mathrm{5}.\mathrm{338} \\ $$
Commented by mr W last updated on 01/Nov/19

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