Question

A piece of wire 44 cm long is cut into two parts and each part is bent to form a square. If the total area
of the two squares is 65 cm2. Find the perimeter of each square.​

Answer 1

Answer:

Let the length of one piece of wire be= x

The length of other piece= 44- x

So the area of squares firmed by these squares are

=(x/4)^2 and {(44-x)/4}^2 respectively.

So (x/4)^2 + {(44-x)/4}^2= 73

Simplifying we get x^2–44x+384=0 Factorising we get (x-32) (x-12)= 0.

So x= 32 or 12.

So the perimetre of square-1= 32. So it's side= 8.

The permetre of square-2=12. So it's side= 3.

So 8^2+ 3^2= 64+ 9= 73.

The side of equilateral ∆-1= 32/3.

So the Area of equilateral∆-1 =√3/4x(32/3)^2 …… Eqn--1

The side of the equilateral ∆-2 =12/3= 4

So the area of the equilateral∆-2 =√3/4 x 4^2. …Eqn--2

So total Area of ∆’s

= √3/4[{32 x 32/9 + 16}

= 56.1954

~ = 56.2 sq.cm

Answer 2

Question :

$\longrightarrow$ A piece of wire is cut into two parts and each part is bent to form a square . If the total area of the two squares is 65cm² , Find the perimeter of each square .

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Answer :

$\:$

${ \fcolorbox{maroon}{teal}{{ \fcolorbox{maroon}{teal}{{ \fcolorbox{maroon}{teal}{{ \fcolorbox{maroon}{teal}{{ \fcolorbox{maroon}{goldenrod}{ \bf{16cm\:and\:28cm}}}}}}}}}}}$

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Length is the measurement of how long a certain object is . Most common units of length are meters , inches , and feet .

A square is a parallelogram having all sides equal . Meaning , unlike a rectangle it's length and width are equal . In a polygon the sum of all the lengths of the sides is called a perimeter .

Basically , the perimeter talks about how long is the boundary enclosing a certain area . So , in a square . . the perimeter is just the total of sides or simply $\mathtt{"4x\: length\: of\: Side"}$

The problem talks about the wire which is 44 cm long and is cut into 2 parts .

we let " x " as the length of the other part and " 44 - x " as the other part .

The lengths will automatically be the perimeter of each square . Since both parts are bend to form a square , then the total perimeter of the two squares is 44cm ( the length of the wire ) .

Given that all sides of the square are equal then the lengths of each sides of the square are :-

• First = $\Large\frac{x}{4}$

$\leadsto$ Area 1 = $\Large\frac{x²}{4}$

• Second = $\Large\frac{44-x}{4}$

$\leadsto$ Area 2 = $\huge{[}$ $\Large\frac{(44-x)}{4}$ $\huge{]}$ ²

The problem further talks about the total area of the two squares which is 65 cm ²

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$\large\mathcal{\underline{๏| substituting\:the \: values :- }}$

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$\large\mathcal{\color{maroon}{A1\:+\:A2\:=\:6cm²}}$

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$\large\bold{\frac{x²}{4}}$ + $\huge{[}$ $\large\bold{\frac{(44-x)}{4}}$ $\huge{]}$² = 65cm²

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$\large\mathcal{\underline{๏| Simplifying\:the\: equation:- }}$

$\huge{(}$ $\large\bold{\frac{(x²)}{16}}$ $\huge{)}$ + $\large\bold{\frac{(1936-88x+x²)}{16}}$ = 65

$\Large\bold{\frac{(x² + 1936 - 88x + x²)}{16}}$

$\mathtt{2(x)²-88x+1936-1040=0}$

$\mathtt{2x²-88x+896=0}$

$\mathtt{x²-44x+448=0}$

$\mathtt\green{x=28,16}$

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Recall that we let " x " as the length of the other part being cut . Thus , one of the perimeter is either 28cm or 16cm .

Solving for the other perimeter , we get :-

$\mathtt\green{using\:x= 28}$

$\mathtt{44-x=44-28=16cm}$

$\mathtt\green{using\:x=16}$

$\mathtt{44-16=28cm}$

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Notice that either of the two values of x give the same values as the other in reverse manner .

Thus , we can say that the perimeter of square are 16 cm and 28 cm .

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