Respuesta:
[tex]a_{1} = 240 \: \: {cm}^{2} \\ a_{2} = 161\pi \: \: {m}^{2} [/tex]
Explicación:
Sea a_1 el área de la primera figura
Sea a_2 el área de la segunda figura
[tex]20 + 2 \alpha = 36 \\ 2 \alpha = 16 \\ \alpha = 8 \\ \\ a_{1} = (10 \: cm \times 20 \: m) + \frac{(8 \: cm \times 10 \: cm)}{2} ) \\ \\ a_{1} = 200 \: {cm}^{2} + \frac{80 \: {cm}^{2} }{2} \\ \\ a_{1} = 200 \: {cm}^{2} + 40 \: {cm}^{2} \\ \\ a_{1} = 240 \: {cm}^{2} [/tex]
[tex] \\ a_{2} = \pi {(15 \: m)}^{2} - \pi {(8 \: m)}^{2} \\ a_{2} = \pi{( {(15 \: m)}^{2} - {(8 \: m}^{2})) } \\ a_{2} = \pi(225 \: {m}^{2} - 64 \: {m}^{2} ) \\ a_{2} = 161\pi \: {m}^{2} [/tex]
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Respuesta:
[tex]a_{1} = 240 \: \: {cm}^{2} \\ a_{2} = 161\pi \: \: {m}^{2} [/tex]
Explicación:
Sea a_1 el área de la primera figura
Sea a_2 el área de la segunda figura
[tex]20 + 2 \alpha = 36 \\ 2 \alpha = 16 \\ \alpha = 8 \\ \\ a_{1} = (10 \: cm \times 20 \: m) + \frac{(8 \: cm \times 10 \: cm)}{2} ) \\ \\ a_{1} = 200 \: {cm}^{2} + \frac{80 \: {cm}^{2} }{2} \\ \\ a_{1} = 200 \: {cm}^{2} + 40 \: {cm}^{2} \\ \\ a_{1} = 240 \: {cm}^{2} [/tex]
[tex] \\ a_{2} = \pi {(15 \: m)}^{2} - \pi {(8 \: m)}^{2} \\ a_{2} = \pi{( {(15 \: m)}^{2} - {(8 \: m}^{2})) } \\ a_{2} = \pi(225 \: {m}^{2} - 64 \: {m}^{2} ) \\ a_{2} = 161\pi \: {m}^{2} [/tex]