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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 sq. cm. Find the dimensions of the smaller square.

(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 14 cm

User Znelson
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1 Answer

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Final answer:

The dimensions of the smaller square, given that the side of the larger square exceeds it by 4 cm and their combined areas equal 400 sq. cm, are found to be 12 cm.

Step-by-step explanation:

The side of a smaller square exceeds the side of another by 4 cm, and the sum of their areas is 400 sq. cm. To find the dimensions of the smaller square, let's denote the side length of the smaller square as s cm. Therefore, the side of the larger square will be s + 4 cm. The area of a square is given by the square of the side length, so the area of the smaller square is s2 and the area of the larger square is (s + 4)2.

The sum of the areas is 400 sq. cm, so we can set up the equation:

s2 + (s + 4)2 = 400

Expanding the second term we get:

s2 + s2 + 8s + 16 = 400

This simplifies to:

2s2 + 8s + 16 = 400

Subtracting 400 from both sides we have:

2s2 + 8s - 384 = 0

Dividing by 2 to simplify:

s2 + 4s - 192 = 0

Factoring the quadratic equation we get:

(s + 16)(s - 12) = 0

Therefore, s could be -16 or 12, but since a side length can't be negative, s must be 12 cm. So the correct answer is (c) 12 cm.

User Sati
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