SOLUTION: A piece of string 90 cm long is cut into two pieces of different lengths. The longer piece of x cm is used to form a rectangle with the width a and length b cm, and the shorter pie
Question 1111133: A piece of string 90 cm long is cut into two pieces of different lengths. The longer piece of x cm is used to form a rectangle with the width a and length b cm, and the shorter piece is used to form a square with a side of length c cm, such that the rectangle and square have the same areas.
a) Prove that and
b) Prove that 0< c <11.25
c) Hence, find that dimensions of the rectangle and square given that they are integers.
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Peimeter of rectangle = 2(a+b)
perimeter =x
x=2(a+b)
(a+b) =x/2 .........(1)
perimeter of square = 4c
4c = 90-x
c = (90-x)/4...............(2)
4c=90-x
x=90-4c
and x = 2(a+b) from (1)
90-4c = 2(a+b)
4c = 90-2(a+b)
c= 22.5 -1/2(a+b)
1/2(a+b) >11.25
therefore c <11.25
a,b,c are integers
c <11.25
areas are equal
c^2 = ab
c is an integer
11^2 = a*b
If c = 11 then a,b will be 11
But a,b are sides of rectangle they cannot be equal
if c= 10
c^2=ab
100= a.b
factors of 100 are 20 & 5
so a = 20 and b = 5,c=10
Perimeter is 50 which is greater than 45 ( so it is the longer part)
or
a =25 and b =4 , c=10
