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Question-100793




Question Number 100793 by bramlex last updated on 28/Jun/20
Commented by bobhans last updated on 28/Jun/20
The perimeter of isosceles ΔMBC = 5+5+6 = 16
$$\mathrm{The}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{isosceles}\:\Delta\mathrm{MBC}\:=\:\mathrm{5}+\mathrm{5}+\mathrm{6}\:=\:\mathrm{16}\: \\ $$
Answered by Dwaipayan Shikari last updated on 28/Jun/20
radius of circle=3 unit  so  (√((7−3)^2 +3^2 ))=5 unit  perimetre of △BMC=5+5+6=16 unit
$${radius}\:{of}\:{circle}=\mathrm{3}\:{unit} \\ $$$${so}\:\:\sqrt{\left(\mathrm{7}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\mathrm{5}\:{unit} \\ $$$${perimetre}\:{of}\:\bigtriangleup\mathrm{BMC}=\mathrm{5}+\mathrm{5}+\mathrm{6}=\mathrm{16}\:{unit} \\ $$

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