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Question

(C) Let the side of the first square be $x\, cm$.Then,the side of the second square is $(x + 4)\, cm$.According to the problem,the sum of their areas

Vedclass · Accepted Answer

$12\, cm$ and $16\, cm$

Vedclass · Answer

(C) Let the side of the first square be $x\, cm$.Then,the side of the second square is $(x + 4)\, cm$.According to the problem,the sum of their areas is $400\, cm^2$:$x^2 + (x + 4)^2 = 400$$x^2 + x^2 + 8x + 16 = 400$$2x^2 + 8x - 384 = 0$Dividing by $2$,we get:$x^2 + 4x - 192 = 0$Factoring the quadratic equation:$x^2 + 16x - 12x - 192 = 0$$x(x + 16) - 12(x + 16) = 0$$(x - 12)(x + 16) = 0$Since the side length cannot be negative,$x = 12$.Thus,the side of the first square is $12\, cm$ and the side of the second square is $12 + 4 = 16\, cm$.

The side of a square exceeds the side of another square by $4\, cm$ and the sum of the areas of the two squares is $400\, cm^2$. The dimensions of the squares are?

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