Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

Commented by mr W last updated on 23/Jun/22

very nice!

$${very}\:{nice}! \\ $$

Commented by Tawa11 last updated on 25/Jun/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com