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53 Determina il perimetro e l'area del quadrilatero \( A B C D \) di vertici \( A(-3,2), B(-2,0), C(2,1), D(1,4) \) [Perimetro \( =3 \sqrt{5}+\sqrt{17}+\sqrt{10} \); 54 Determina il perimetro e l'area del quadrilatero \( A B C D \) di vertici \( A(-2,3), B(2,0), C(5,2), D(0,6) \)

Ask by Cook Peters. in Italy
Mar 09,2025

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**Problema 53:** - **Perimetro:** \( 3\sqrt{5} + \sqrt{17} + \sqrt{10} \) - **Area:** \( \frac{23}{2} \) **Problema 54:** - **Perimetro:** \( 5 + 2\sqrt{13} + \sqrt{41} \) - **Area:** \( 20 \)

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Per calcolare il perimetro del quadrilatero \( ABCD \) con vertici \( A(-2,3), B(2,0), C(5,2), D(0,6) \), si devono trovare le lunghezze dei lati utilizzando la formula della distanza tra due punti. I lati sono \( AB \), \( BC \), \( CD \) e \( DA \). Dopo aver calcolato le distanze: - \( AB = \sqrt{(2 - (-2))^2 + (0 - 3)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{25} = 5 \) - \( BC = \sqrt{(5 - 2)^2 + (2 - 0)^2} = \sqrt{3^2 + 2^2} = \sqrt{13} \) - \( CD = \sqrt{(0 - 5)^2 + (6 - 2)^2} = \sqrt{(-5)^2 + 4^2} = \sqrt{41} \) - \( DA = \sqrt{(-2 - 0)^2 + (3 - 6)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{13} \) Quindi, il perimetro sarà \( P = AB + BC + CD + DA = 5 + \sqrt{13} + \sqrt{41} + \sqrt{13} = 5 + 2\sqrt{13} + \sqrt{41} \). Per calcolare l'area del quadrilatero, si può utilizzare la formula di Shoelace: \[ Area = \frac{1}{2} | x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) | \] Sostituendo i vertici, l'area risulta: \[ Area = \frac{1}{2} | (-2)(0) + 2(2) + 5(6) + 0(3) - (3(2) + 0(5) + 2(0) + 6(-2)) | \] Calcolando, si ottiene l'area corretta. Spero che questi calcoli ti aiutino! Divertiti con la geometria!

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