Problema

BC+10BC/13 = 23BC/13 = 92 cm

lato BC = lo = 92/23*13 = 52 cm 

base AB = b = 92-52 = 40 cm 

altezza h = √lo^2-(b/2)^2 = 26√26^2-10^2 = 48 cm

perimetro 2p = 2*52+40 = 144 cm

area a = b*h/2 = 48*20 = 960 cm^2

92/(10+13)=4    b=4*10=40   l=13*4=52    h=V 52^2-20^2=48   A=48*40/2=960cm2

2p=52*2+40=144cm

160) Triangolo isoscele

Somma (92 cm) e rapporto (10/13) tra base e lato obliquo, quindi:

base $\small b= \dfrac{92}{10+13}×10 = \dfrac{\cancel{92}^4}{\cancel{23}_1}×10 = 4×10 = 40\,cm;$

ciascun lato obliquo $\small l= \dfrac{92}{10+13}×13 = \dfrac{\cancel{92}^4}{\cancel{23}_1}×13 = 4×13 = 52\,cm;$

altezza $\small h= \sqrt{l^2-\left(\dfrac{b}{2}\right)^2} = \sqrt{52^2-\left(\dfrac{40}{2}\right)^2} = \sqrt{52^2-20^2} = 48\,cm$ (teorema di Pitagora);

perimetro $\small 2p= b+2l = 40+2×52 = 40+104 = 144\,cm;$

area $\small A= \dfrac{b×h}{2} = \dfrac{\cancel{40}^{20}×48}{\cancel2_1} = 20×48 = 960\,cm^2.$