Math  /  Geometry

Questionle Rerimelice est isule maximom 55om5 \mathrm{5om} Largiar =x=x tong bres =4x=4 x sa largeur es isale an avadruple de salargan Calculer les mesue possiale des âtes du rectangie 2 e calculer ĺaile du reciangle. P=2(L+l2(4x+x)2302(5x)250l=x=15L=4x=60{A=15×60=900 cm=900 cm Pra 900×0+x70,25]m25 Longuar =4x=4×25L=100 m & [0,25)l=rL=4YL=20 mx=20=llˉ=3 mL=40 mC=4x=4×20=801=80×20=1600l=16 ml=x=10l=4x=40prx×1 m=sos1 solution \begin{array}{l} P=2 \cdot(L+l \\ 2 \cdot(4 x+x) \leq 230 \\ 2(5 x) \leq 250 \\ l=x=15 \text {. } \\ L=4 x=60 \\ \left\{\begin{aligned} A & =15 \times 60 \\ & =900 \mathrm{~cm} \end{aligned}\right. \\ =900 \mathrm{~cm} \\ \text { Pra } 900 \times 0 \\ +\infty \\ x \in 70,25] \mathrm{m} \\ 25 \\ \text { Longuar }=4 x=4 \times 25 \\ L=\leqslant 100 \mathrm{~m} \\ \text { \& } \in[0,25) \\ l=r \\ L=4 Y \\ L=20 \mathrm{~m} \\ x=20=l \\ \bar{l}=3 \mathrm{~m} \\ L=40 \mathrm{~m} \\ C=4 x=4 \times 20=80 \\ 1=80 \times 20=1600 \\ \begin{array}{ll} l=16 \mathrm{~m} & l=x=10 \\ l=4 x=40 \end{array} \\ \operatorname{pr} x \times 1 \mathrm{~m}^{\prime}=\operatorname{sos} \\ 1 \text { solution } \end{array} L=4x=40 mL=4x=60A1=10×40 m2A=15×60A=20×80=160\begin{array}{l} L=4 x=40 \mathrm{~m} \\ L=4 x=60 \\ A_{1}=10 \times 40 \mathrm{~m}^{2} \\ A=15 \times 60 \\ A=20 \times 80 \\ =160 \end{array}

Studdy Solution
Calculate the area of the rectangle:
A=l×L A = l \times L A=23×92 A = 23 \times 92 A=2116 square meters A = 2116 \text{ square meters}
The possible dimensions of the rectangle are a width of 23 23 meters and a length of 92 92 meters, with an area of:
2116 square meters \boxed{2116 \text{ square meters}}

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