If the square has side s, then
4s = 2(x+4 + x)
s^2 + x(x+4) = 94
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Solve the following word problem.
The length and width of a rectangle are (x + 4) cm and x cm. Respectively, a square has the same perimeter as this rectangle. If you add the areas of the rectangle and the square, the result is 94cm^2. Solve for x.
Note: Write your answer as x = ____ and x = ____
2 answers
Area rectangle = x (x + 4) = x^2 + 4x
Perimeter of rectangle = 2 (x + x + 4)= 4x + 8
Perimeter of square = perimeter of rectangle = 4x + 8
Therefore length of square side = .25 (4x + 8) = x + 2
and area of square = (x + 2)^2 = x^2 + 4x + 4
Area of square + area of rectangle = 94 cm^2
Therefore:
x^2 + 4x + x^2 + 4x + 4 = 94
Or, after some rearrangement of terms:
x^2 + 4x - 45 = 0
Use quadratic theorum to obtain:
x = 5 cm or -9 cm
Choose x = 5 cm since a negative area is not physically real.
QED
Perimeter of rectangle = 2 (x + x + 4)= 4x + 8
Perimeter of square = perimeter of rectangle = 4x + 8
Therefore length of square side = .25 (4x + 8) = x + 2
and area of square = (x + 2)^2 = x^2 + 4x + 4
Area of square + area of rectangle = 94 cm^2
Therefore:
x^2 + 4x + x^2 + 4x + 4 = 94
Or, after some rearrangement of terms:
x^2 + 4x - 45 = 0
Use quadratic theorum to obtain:
x = 5 cm or -9 cm
Choose x = 5 cm since a negative area is not physically real.
QED
1 answer