Question

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 65 in squared . Find the lengths of the sides of the two squares.

Answer 1

Answer:

$L_1 = 4$

And $L_2 = 3 +4 = 7$

Step-by-step explanation:

For this case we assume that for the first square we have the following dimensions:

$A_1 = L^2_1$

And we know that:

$L_2 = 3 + L_1$

And the area for the second square would be:

$A_2 = L^2_2 = (3+L_1)^2 = 9 + 6L_1 + L^2_1$

And we know that the sum of areas is 65 so then we have this:

$A_1 + A_2 = 65$

And replacing we got:

$L^2_1 + 9 + 6L_1 + L^2_1 = 65$

$2L^2_1 +6L_1 - 56=0$

We can divide the last expression by 2 and we got:

$L^2_1 + 3L_1 -28=0$

And we can factorize the last expression like this:

$(L_1 + 7) (L_1 -4) =0$

And we have two solutions for $L_1$ and we got:

$L_1 = 4, L_1 = -7$

Since the length can't be negative we have this:

$L_1 = 4$

And $L_2 = 3 +4 = 7$

Answer 2

Answer:

4 inch and 7 inch

Step-by-step explanation:

Let the length of the smaller square=l

Since the length of each side of a square is 3 in. more than the length of each side of a smaller square,

Length of the bigger square=l+3

Area of Smaller Square=l²

Area of Larger Square=(l+3)²

The sum of their areas is 65 inch squared

Therefore: l²+(l+3)²=65

l²+(l+3)(l+3)=65

l²+l²+3l+3l+9=65

2l²+6l+9-65=0

2l²+6l-56=0

2l²+14l-8l-56=0

2l(l+7)-8(l+7)=0

(2l-8)(l+7)=0

2l-8=0 or l+7=0

l=4 or -7

l= 4 inch

The length of the larger square is (4 +3) inch =7 inch