Esercizi svolti per il libro Matematica.blu: Algebra, Geometria, Statistica, Vol. 1
Soluzione
BH=b+b=2bBH=b+b=2bBH=b+b=2b; AB=AH+BH=a+2bAB=AH+BH=a+2bAB=AH+BH=a+2b; Area del triangolo ABC=AB⋅CH2=(a+2b)⋅a2ABC=\dfrac{AB\cdot CH}{2} = \dfrac{(a+2b)\cdot a}{2}ABC=2AB⋅CH=2(a+2b)⋅a === a2+2ab2\dfrac{a^2+2ab}{2}2a2+2ab.
Area del triangolo ACH=AH⋅CH2=a⋅a2=a22ACH=\dfrac{AH\cdot CH}{2} = \dfrac{a\cdot a}{2}=\dfrac{a^2}{2}ACH=2AH⋅CH=2a⋅a=2a2.
Area del triangolo CHB=BH⋅CH2=2b⋅a2=2ab2=abCHB=\dfrac{BH\cdot CH}{2} = \dfrac{2b \cdot a}{2} = \dfrac{\cancel2ab}{\cancel2} = abCHB=2BH⋅CH=22b⋅a=22ab=ab.