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The length of each side of a square is 3 inches less than the length of a side of an equilateral triangle. If the perimeters of the two figures are the same, what is the measure of the sides of each figure

User Torkel
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Answer:

Explanation:

The length of each side of a square is 3 inches more than the length of each side of a smaller square.

Let's let the smaller square side be variable X.

One side of the larger square is X+3.

The sum of the areas of the squares is 185 inches2

The area if a square is the square of the sides (x2) for the smaller one and (x+3)2 for the larger one.

The area of the larger one would be x2+6x+9. Now we add the smaller square

x2+6x+9+x2=185

Combine the like terms and you have 2x2+6x+9=185

Subtract 9 from both sides you get 2x2+6x=176

To write in standard form we need to subtract 176 from both sides so the equation will equal zero.

2x2+6x-176=0

We can factor out a 2:

x2+3x+88=0

We can factor and then solve:

x2+11x-8x-88=0

X(x+11)-8(x+11)=0

(x-8)(x+11)=0

(x-8)=0 or (x+11)=0

X=8 or x=-11

We can drop the negative number. So the side of the smaller square is 8 inches and the side of the larger square is 11 inches.

Now lets check it: 8*8=64 and 11*11 is 121. 64+121=185.

User Satish Dhiman
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