Halve grote as = 5     Halve kleine as = 4
    A(5,0)           B(0,4)
rico AB is \(\frac {4-0} {0-5}=-\,\frac45 \)   (vergelijking van AB hebben we niet nodig)
rico BC is \(\frac54\) ⇒ vergelijking van BC is   \(y-4=\frac54(x-0)=\frac54 x\)
de rechte BC snijdt de x-as in \(-\,\frac{16}5\)
Lengte van de rechthoekszijden van ΔABC :
\(\small|AB|=\sqrt{4^2+5^2}=\sqrt{16+25}=\sqrt{41}\)
\(\small|BC|=\sqrt{4^2+(-\,\frac{16}5)^2} = \sqrt{16+\frac{16^2}{25}}=4.\sqrt{1+\frac{16}{25}}=4\sqrt{\frac{41}{25}}=\frac45.\sqrt{41}\)
De oppervlakte van ΔABC is \(\frac{|AB|.|BC|}{2}=\frac12\sqrt{41}.\frac45\sqrt{41}=\frac{4.41}{10}=16,4 \)