Question

A Piece of wire 44 cm long is cut into two parts. Each is bent to form a square. Given that the total area of two square is 65 cm², find the perimeter of each square? (Synthesis base)​

Answer 1

Step-by-step explanation:

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

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