A piece of wire of length 44 cm is cut into two parts. Each part is bent to form the four sides of a square?

Question

a) If the length of a side of one square is x cm, what is the length of the side of the other square?

b) the sum of the areas is 65 cm^2. Find x

Answer 1

Suppose the perimeter of one square is X
the perimeter of onother square is 44-X

So, the side of 1 square is X/4 and
So, the side of another square is 44-X/4
So, the sum of both squares is,
(X/4)^2+(44-X/4)^2
=65 cm^2
so,
X^2/16 + (44^2 - 2*44*X + X^2)/16=65
X^2 +44^2 - 88*X + X^2=65*16(Multiplying both sides by 16)
2*X^2- 88*X [phone]=0
2*X^2- 88*X + 896=0
X^2- 44*X + 448=0(DIVIDING BOTH SIDES BY 2)
NOW,
Roots are:
comparing eq. with ax^2+b*x+c
a=1,b= -44 , c=448
X= {-b+/-sqrt(b^2-4*a*c)/2*a}
X=44+sqrt[(-44)^2-4*1*448]/2*1& X=44-sqrt[(-44)^2-4*1*448]/2*1
X=[44+sqrt(1936-1792)]/2 & X=[44-sqrt(1936-1792)]/2
X=[44+SQRT144]/2 & X=[44-SQRT144]/2
X=(44+12)/2 & X=(44-12)/2
X=56/2 & 32/2
X=28 & X=16
Since X is a perimeter,
side of one square
=28/4
=7 cm
side of another square is (44-X)/4
=(44-28)/4
=16/4
=4 cm.
or,side of another square is,
16/4
=4 cm

Answer 2

If a and b are the lengths of the two parts,
a + b = 44,
Arbitrarily set a = 4x
4x + b = 44
y = side length of other square
4x + 4y = 44,
a) y = 11 - x

Sum of areas = 65 cm^2,
= x^2 + y^2 = x^2 + x^2 - 22x + 121
x^2 - 11x + 28 = 0

b) x = 7 cm or 4 cm, (the answer is not unique)

Answer 3

a)

part A = X cm
part B = y = 44-X cm

x = X/4

y = (44-X)/4 = 11 - X/4 = 11 - x

the length of the side of the second square is : y = 11 - x

b)

area of the first square is : A = x²
area of the second square is : B = y² = (11 - x)² = 121 + x² - 22x

sum A + B = 65

x² + 121 + x² -22x = 65

2x² - 22x +56 = 0

then : x² - 11x + 28 = 0

Delta = (-11)² - 4*(1*28) = 121 - 112 = 9

sol_1 = (11 - sqrt(9))/2 = (11 - 3)/2 = 4
sol_2 = (11 + sqrt(9))/2 = (11 + 3)/2 = 7

in fact, this is the same solution, in the first case (x=4) the first square is smaller than the second (side y = 7) ... in the other case (x=7) the first square is bigger than the second (y = 4)

Answer 4

Start by setting up your two equations, the length of the side of one square being X and the length of a side of the other square being Y. We know that

The area of the two squares is 65cm^2 or
X^2 + Y^2 = 65

and

The total circumferences of the 2 squares is 44 or
4X + 4Y = 44

That should get you close enough to solve it without doing your homework ;-)

Answer 5

a) perimeter of Sq1=4x
perimeter of Sq2=44-4x
length of one side of Sq2=(44-4x)/4=11-x

b)Area 1=x^2
area2= (11-x)^2= 121-22x+x^2
total area=x^2+121+x^2-22x
2x^2-22x+121=65
2x^2-22x+56=0
x^2-11x+28=0
x^2-7x-4x+28=0
x(x-7)-4(x-7)=0
x=7 or x=4
smaller square x=4 larger square x=7

Answer 6

enable a = the size of an factor of the smaller component of the twine and enable b= an factor of the greater desirable twine sq.. 4a+4b=100cm a+b=25cm a=25cm-b a^2+b^2=425cm^2 (25cm-b)^2+b^2=425cm^2 625cm^2-50b+2b^2=425cm^2 2b^2-50b-200cm=0 b^2-25b-100cm=0 Now only sparkling up with the quadratic formula to get b=20cm, 5cm. considering the fact that we are in seek of the greater desirable section, that is 20cm.