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): You can put this solution on YOUR website!

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.

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

            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.

Solved.

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

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....

(1)^2 + (1+3)^2 = 1+16 = 17 nope...
(2)^2+(2+3)^2 = 4+25 = 29 nope...
(3)^2+(3+3)^2 = 9+36 = 45 YES!!

ANSWER: The side lengths ar 3 and 6.

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