Respuestas
Respuesta dada por:
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} 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}](https://tex.z-dn.net/?f=P_%7B1%7D%3D2.80%2B13%2B2.80%2B13%5C%5CP_%7B1%7D%3D2%282.80%29%2B2%2813%29%3D31.6%5C%2Ccm%5C%5CA_%7B1%7D%3Dbh%3D%282.80%5C%2Ccm%29%2813%5C%2Ccm%29%3D36.4%5C%2Ccm%5E%7B2%7D)
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} 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}](https://tex.z-dn.net/?f=a%5E%7B2%7D%2Bb%5E%7B2%7D%3Dc%5E%7B2%7D%5C%5Ca%3D3.5%5C%2Ccm%5C%5Cb%3D4.8%5C%2Ccm%5C%5Cc%5E%7B2%7D%3Da%5E%7B2%7D%2Bb%5E%7B2%7D%5C%5Cc%3D%5Csqrt%7Ba%5E%7B2%7D%2Bb%5E%7B2%7D%7D%5C%5Cc%3D%5Csqrt%7B%283.5%5C%2Ccm%29%5E%7B2%7D%2B%284.8%5C%2Ccm%29%5E%7B2%7D%7D%5C%5Cc%3D5.9%5C%2Ccm%5C%5CP_%7B2%7D%3D3.5%5C%2Ccm%2B4.8%5C%2Ccm%2B5.9%5C%2Ccm%3D14.2%5C%2Ccm%5C%5CA_%7B2%7D%3D%5Cfrac%7Bbh%7D%7B2%7D%5C%5CA_%7B2%7D%3D%5Cfrac%7B%283.5%5C%2Ccm%29%284.8%5C%2Ccm%29%7D%7B2%7D%3D8.4%5C%2Ccm%5E%7B2%7D%5C%5CP_%7B3%7D%3D5l%3D5%2810.2%5C%2Ccm%29%3D51%5C%2Ccm%5C%5CA_%7B3%7D%3D%5Cfrac%7BPa%7D%7B2%7D%3D%5Cfrac%7B%2851%5C%2Ccm%29%288.2%5C%2Ccm%29%7D%7B2%7D%5C%5CA_%7B3%7D%3D209%5C%2Ccm%5E%7B2%7D%5C%5CP_%7B4%7D%3D180%5C%2Ccm%5C%5CA_%7B4%7D%3D%5Cfrac%7BPa%7D%7B2%7D%3D%5Cfrac%7B%28180%5C%2Ccm%29%2827.16%5C%2Ccm%29%7D%7B2%7D%5C%5CA_%7B4%7D%3D2444.4%5C%2Ccm%5E%7B2%7D%5C%5CP_%7B5%7D%3D8l%5C%5Cl%3D17.3%5C%2Ccm%5C%5CP_%7B5%7D%3D8%2817.3%5C%2Ccm%29%3D138.4%5C%2Ccm%5C%5CA_%7B5%7D%3D%5Cfrac%7BPa%7D%7B2%7D%3D%5Cfrac%7B%28138.4%5C%2Ccm%29%28266.2%5C%2Ccm%29%7D%7B2%7D%5C%5CA_%7B5%7D%3D18421%5C%2Ccm%5E%7B2%7D)
Saludos
Para.encontrar el otro lado del triángulo aplicamos el teorema de Pitagoras y nos queda:
Saludos
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