Weber Powers
10/03/2023 · Primary School
12. [s punti] In un trapezio isoscele la diffe- renza delle due basi misura 72 cm e l'altezza è \( \frac{2}{3} \) di questa differenza. Sapendo che l'area del trapezio è \( 3648 \mathrm{~cm}^{2} \) calcola l'area di un rettangolo avente le dimensioni lunghe quanto le due basi del trapezio. \( \begin{array}{lll}\text { a. } 4480 \mathrm{~cm}^{2} & \text { b. } 1600 \mathrm{~cm}^{2} & \text { c. } 2240 \mathrm{~cm}^{2}\end{array} \)
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L'area del rettangolo è 4480 cm².
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