Question-172062

Question Number 172062 by cortano1 last updated on 23/Jun/22

Answered by kapoorshah last updated on 23/Jun/22

s = ((10 + 14 + 16)/2) = 20 r = (√(((20 − 10)(20 − 14)(20 − 16))/(20))) = 2(√3) a + b = 16 b + c = 10 c + a = 14 ⇒ a = 10, b = 6, c = 4 x = (√(10^2 + (2(√3))^2 )) = 4(√7) y = (√(4^2 + (2(√3))^2 )) = 2(√7) z = (√(6^2 + (2(√3))^2 )) = 4(√3) x + y + z = 6(√7) + 4(√3)

$${s}\:=\:\frac{\mathrm{10}\:+\:\mathrm{14}\:+\:\mathrm{16}}{\mathrm{2}} \\ $$$$\:\:\:=\:\mathrm{20} \\ $$$$ \\ $$$${r}\:=\:\sqrt{\frac{\left(\mathrm{20}\:−\:\mathrm{10}\right)\left(\mathrm{20}\:−\:\mathrm{14}\right)\left(\mathrm{20}\:−\:\mathrm{16}\right)}{\mathrm{20}}} \\ $$$$\:\:\:=\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${a}\:+\:{b}\:=\:\mathrm{16} \\ $$$${b}\:+\:{c}\:=\:\mathrm{10} \\ $$$${c}\:+\:{a}\:=\:\mathrm{14} \\ $$$$ \\ $$$$\Rightarrow\:{a}\:=\:\mathrm{10},\:{b}\:=\:\mathrm{6},\:{c}\:=\:\mathrm{4} \\ $$$$ \\ $$$${x}\:=\:\sqrt{\mathrm{10}^{\mathrm{2}} \:+\:\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\: \\ $$$$\:\:\:\:=\:\mathrm{4}\sqrt{\mathrm{7}} \\ $$$$ \\ $$$${y}\:=\:\sqrt{\mathrm{4}^{\mathrm{2}} \:+\:\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:=\:\mathrm{2}\sqrt{\mathrm{7}} \\ $$$$ \\ $$$${z}\:=\:\sqrt{\mathrm{6}^{\mathrm{2}} \:+\:\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\:\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${x}\:+\:{y}\:+\:{z}\:=\:\mathrm{6}\sqrt{\mathrm{7}}\:+\:\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by kapoorshah last updated on 23/Jun/22