A piece of string 90 cm long is cut into two pieces of different lengths. The longer piece of x(cm) is used for a rectangle with width a cm and length b cm, and

Question

A piece of string 90 cm long is cut into two pieces of different lengths. The longer piece of x(cm) is used for a rectangle with width a cm and length b cm, and the shorter piece is used to form a square with side c cm, such that the rectangle and the square have the same areas. 

 

a. Prove that a+b=x/2 and c=(90-x)/4

 

b. Prove that 0<c<11.25

 

c. Hence, find the dimensions of the rectangle and the square given that they are integers. 

Answer 1

A piece of string 90 cm long is cut into two pieces of different lengths. The longer piece of x(cm) is used for a rectangle with width a cm and length b cm, and the shorter piece is used to form a square with side c cm, such that the rectangle and the square have the same areas.

a. Prove that a+b=x/2 and c=(90-x)/4

b. Prove that 0<c<11.25

c. Hence, find the dimensions of the rectangle and the square given that they are integers.

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A.   x=2(a+b) i.e. perimeter of rectangle is the X

    or,  a+b=x/2

  (90-x) is the perimeter of square, or:  c=(90-x)/4

 

B.  maximum area with fixed perimeter corresponds to square.

     Let's assume  c>x  : then  c^2>11.25^2

   Same time, x<45 and maximum of area formed from perimeter x is (x/4)^2< 11.25^2

   This means, those two areas are different.

    Although, assumption c>x contradicts to problem's conditions: it says "The longer piece of x(cm) ..."

 

C.

 

    c=10  ;  a+b=25  ;  a=5 , b=20 ; all in centimeters