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Question-203009




Question Number 203009 by cortano12 last updated on 07/Jan/24
Answered by som(math1967) last updated on 07/Jan/24
Perimeter=80  AD=AG+GD=8+8=16unit  AB=((80)/2) −16=24unit  FB=24−8=16  FQ=16−R  OP=8+R  (8+R)^2 =(8−R)^2 +(16−R)^2   32R=256−32R+R^2   R^2 −64R+256=0   R=((64−(√(64^2 −4×256)))/2)  =((64−(√((64+32)(64−32))))/2)  =((64−(√(96×32)))/2)  =((64−32(√3))/2)  =16(2−(√3))  =4.28 unit (approx)
$${Perimeter}=\mathrm{80} \\ $$$${AD}={AG}+{GD}=\mathrm{8}+\mathrm{8}=\mathrm{16}{unit} \\ $$$${AB}=\frac{\mathrm{80}}{\mathrm{2}}\:−\mathrm{16}=\mathrm{24}{unit} \\ $$$${FB}=\mathrm{24}−\mathrm{8}=\mathrm{16} \\ $$$${FQ}=\mathrm{16}−{R} \\ $$$${OP}=\mathrm{8}+{R} \\ $$$$\left(\mathrm{8}+{R}\right)^{\mathrm{2}} =\left(\mathrm{8}−{R}\right)^{\mathrm{2}} +\left(\mathrm{16}−{R}\right)^{\mathrm{2}} \\ $$$$\mathrm{32}{R}=\mathrm{256}−\mathrm{32}{R}+{R}^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} −\mathrm{64}{R}+\mathrm{256}=\mathrm{0} \\ $$$$\:{R}=\frac{\mathrm{64}−\sqrt{\mathrm{64}^{\mathrm{2}} −\mathrm{4}×\mathrm{256}}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{64}−\sqrt{\left(\mathrm{64}+\mathrm{32}\right)\left(\mathrm{64}−\mathrm{32}\right)}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{64}−\sqrt{\mathrm{96}×\mathrm{32}}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{64}−\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\mathrm{16}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$$=\mathrm{4}.\mathrm{28}\:{unit}\:\left({approx}\right) \\ $$

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