SOLUTION: The length of each side of a square is 3 in. more than the length of each side of a smaller square. The sum of the areas of the squares is 45 in squared . Find the lengths of the s
Question 1151211: The length of each side of a square is 3 in. more than the length of each side of a smaller square. The sum of the areas of the squares is 45 in squared . Find the lengths of the sides of the two squares. Found 3 solutions by jim_thompson5910, ikleyn, greenestamps:Answer by jim_thompson5910(35256) (Show Source):
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Square A has side length x+3
Square B has side length x
Square A has a side length 3 units larger compared to the side length of square B.
Negative side lengths are not possible, so x > 0.
Compute the area of square A
area of square A = (side length of square A)^2
area of square A = (x+3)^2
area of square A = (x+3)(x+3)
area of square A = x(x+3)+3(x+3)
area of square A = x^2+3x+3x+9
area of square A = x^2+6x+9
You could use the FOIL rule to expand out (x+3)^2, but I think the distributive property is more versatile.
Compute the area of square B
area of square B = (side length of square B)^2
area of square B = x^2
Add up the two areas
total area = (area of square A)+(area of square B)
total area = (x^2+6x+9)+(x^2)
total area = 2x^2+6x+9
Set this total area equal to 45 and solve for x
2x^2+6x+9 = 45
2x^2+6x+9-45 = 0
2x^2+6x-36 = 0
2(x^2+3x-18) = 0
x^2+3x-18 = 0/2
x^2+3x-18 = 0
(x+6)(x-3) = 0
x+6 = 0 or x-3 = 0
x = -6 or x = 3
Since x > 0, this means we only consider x = 3 as the solution.
If x = 3, then x+3 = 3+3 = 6.
Square A has a side length of 6 inches and an area of 36 square inches.
Square B has a side length of 3 inches and an area of 9 square inches.
The total area is 36+9 = 45 square inches.
The answer checks out.
I will show you how to solve this problem quickly and mentally.
From the condition, you have this system of 2 equations
x - y = 3 (1)
x^2 + y^2 = 45. (2)
Square equation (1); keep equation (2) as is.
x^2 - 2x + y^2 = 9 (1')
x^2 + y^2 = 45 (2')
Now subtract equation (1') from equation (1'). You will get
2xy = 45 - 9 = 36, or
xy = 36/2 = 18.
Thus you have two numbers x and y with the difference of 3 and the product of 18.
3 seconds is enough to guess (to recognize) these numbers, 3 and 6.
A quick mental solution doesn't require any algebra.
The sum of the areas is a whole number; the difference in the side lengths is a whole number. That virtually assures that both side lengths are whole numbers.
The sum of the areas is a small number; so solve using (organized) trial and error....
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