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[attach]108295[/attach]

remanzini_rinaldo · Accepted Answer

[attach]108943[/attach] 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

pier_effe · Answer

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

gramor · Answer

160) Triangolo isoscele

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

base $b = \frac{92}{10 + 13} \times 10 = \frac{92^{4}}{23_{1}} \times 10 = 4 \times 10 = 40 c m;$

ciascun lato obliquo $l = \frac{92}{10 + 13} \times 13 = \frac{92^{4}}{23_{1}} \times 13 = 4 \times 13 = 52 c m;$

altezza $h = \sqrt{l^{2} - {(\frac{b}{2})}^{2}} = \sqrt{52^{2} - {(\frac{40}{2})}^{2}} = \sqrt{52^{2} - 20^{2}} = 48 c m$ (teorema di Pitagora);

perimetro $2 p = b + 2 l = 40 + 2 \times 52 = 40 + 104 = 144 c m;$

area $A = \frac{b \times h}{2} = \frac{40^{20} \times 48}{2_{1}} = 20 \times 48 = 960 c m^{2} .$

gramor

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18/03/2025 15:23

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