Respuesta :
Answer:
[tex] L_1 = 4[/tex]
And [tex] L_2 = 3 +4 = 7[/tex]
Step-by-step explanation:
For this case we assume that for the first square we have the following dimensions:
[tex] A_1 = L^2_1[/tex]
And we know that:
[tex] L_2 = 3 + L_1[/tex]
And the area for the second square would be:
[tex] A_2 = L^2_2 = (3+L_1)^2 = 9 + 6L_1 + L^2_1[/tex]
And we know that the sum of areas is 65 so then we have this:
[tex] A_1 + A_2 = 65[/tex]
And replacing we got:
[tex] L^2_1 + 9 + 6L_1 + L^2_1 = 65[/tex]
[tex] 2L^2_1 +6L_1 - 56=0[/tex]
We can divide the last expression by 2 and we got:
[tex] L^2_1 + 3L_1 -28=0[/tex]
And we can factorize the last expression like this:
[tex] (L_1 + 7) (L_1 -4) =0[/tex]
And we have two solutions for [tex] L_1[/tex] and we got:
[tex] L_1 = 4, L_1 = -7[/tex]
Since the length can't be negative we have this:
[tex] L_1 = 4[/tex]
And [tex] L_2 = 3 +4 = 7[/tex]
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
