Question

ayuda por favor es de mate please

Answer 1

Bueno si copiastes bien,entonces el perímetro y área de cada figura es:
$P_{1}=2.80+13+2.80+13\\P_{1}=2(2.80)+2(13)=31.6\,cm\\A_{1}=bh=(2.80\,cm)(13\,cm)=36.4\,cm^{2}$
Para.encontrar el otro lado del triángulo aplicamos el teorema de Pitagoras y nos queda:
$a^{2}+b^{2}=c^{2}\\a=3.5\,cm\\b=4.8\,cm\\c^{2}=a^{2}+b^{2}\\c=\sqrt{a^{2}+b^{2}}\\c=\sqrt{(3.5\,cm)^{2}+(4.8\,cm)^{2}}\\c=5.9\,cm\\P_{2}=3.5\,cm+4.8\,cm+5.9\,cm=14.2\,cm\\A_{2}=\frac{bh}{2}\\A_{2}=\frac{(3.5\,cm)(4.8\,cm)}{2}=8.4\,cm^{2}\\P_{3}=5l=5(10.2\,cm)=51\,cm\\A_{3}=\frac{Pa}{2}=\frac{(51\,cm)(8.2\,cm)}{2}\\A_{3}=209\,cm^{2}\\P_{4}=180\,cm\\A_{4}=\frac{Pa}{2}=\frac{(180\,cm)(27.16\,cm)}{2}\\A_{4}=2444.4\,cm^{2}\\P_{5}=8l\\l=17.3\,cm\\P_{5}=8(17.3\,cm)=138.4\,cm\\A_{5}=\frac{Pa}{2}=\frac{(138.4\,cm)(266.2\,cm)}{2}\\A_{5}=18421\,cm^{2}$
Saludos