Bonjour,
1 KC = KH + HC
[tex]KC=\dfrac{15}{21}+\dfrac{5}{14}\\\\KC=\dfrac{3\times5}{3\times7}+\dfrac{5}{14}\\\\KC=\dfrac{5}{7}+\dfrac{5}{14}\\\\KC=\dfrac{2\times5}{2\times7}+\dfrac{5}{14}\\\\KC=\dfrac{10}{14}+\dfrac{5}{14}\\\\KC=\dfrac{15}{14}[/tex]
2) Aire KOC = (1/2) * KC * OH
[tex]Aire\ KOC = \dfrac{1}{2}\times \dfrac{15}{14}\times\dfrac{14}{25}\\\\Aire\ KOC = \dfrac{1}{2}\times \dfrac{15}{25}\\\\Aire\ KOC = \dfrac{1}{2}\times \dfrac{5\times3}{5\times5}\\\\Aire\ KOC = \dfrac{1}{2}\times \dfrac{3}{5}\\\\Aire\ KOC = \dfrac{3}{10}[/tex]
3) Le triangle KHO est rectangle en H.
Par Pythagore, nous avons :
[tex]OK^2=KH^2+HO^2\\\\OK^2=(\dfrac{15}{21})^2+ (\dfrac{14}{25})^2\\\\OK^2 = (\dfrac{5}{7})^2+ (\dfrac{14}{25})^2\\\\OK^2= \dfrac{25}{49}+ \dfrac{196}{625}\\\\OK^2=\dfrac{15625}{30625}+\dfrac{9604}{30625}\\\\OK^2=\dfrac{25229}{30625}\\\\OK=\sqrt{\dfrac{25229}{30625}}\approx 0,9[/tex]
