A wire with length of 52 cm is cut into two unequal lengths. Each par is bent to form a square. If the sum of the area of the two squares is 97 sq. cm., what is the area of the smaller square? A. 16 C. 64 B. 81 D. 49 Solution Click here to… Jhun Vert Fri, 07/05/2024 - 06:10 Solution Click here to expand or collapse this section $A_1 + A_2 = 97$ $\left( \dfrac{x}{4} \right)^2 + \left( \dfrac{52 - x}{4} \right)^2 = 97$ $\dfrac{x^2}{16} + \dfrac{2704 - 104x + x^2}{16} = 97$ $x^2 + (2704 - 104x + x^2) = 1552$ $2x^2 - 104x + 1152 = 0$ $x^2 - 52x + 576 = 0$ $(x - 16)(x - 36) = 0$ $x = 16 ~ \text{and} ~ 36$ $\text{Area of smaller square} = \left( \dfrac{16}{4} \right)^2$ $\text{Area of smaller square} = 16 ~ \text{cm}^2$ Log in or register to post comments Log in or register to post comments
Solution Click here to… Jhun Vert Fri, 07/05/2024 - 06:10 Solution Click here to expand or collapse this section $A_1 + A_2 = 97$ $\left( \dfrac{x}{4} \right)^2 + \left( \dfrac{52 - x}{4} \right)^2 = 97$ $\dfrac{x^2}{16} + \dfrac{2704 - 104x + x^2}{16} = 97$ $x^2 + (2704 - 104x + x^2) = 1552$ $2x^2 - 104x + 1152 = 0$ $x^2 - 52x + 576 = 0$ $(x - 16)(x - 36) = 0$ $x = 16 ~ \text{and} ~ 36$ $\text{Area of smaller square} = \left( \dfrac{16}{4} \right)^2$ $\text{Area of smaller square} = 16 ~ \text{cm}^2$ Log in or register to post comments
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