Question

What are the dimensions of a rectangle that has a perimeter of 74 units and an area of 102 square units

Answer 1

let the dimensions i.e. length and width of the rectangle be a and b respectively
Thus
Area of rectangle =ab
i.e. 102 =ab
or ab= 102--------(1
perimeter of rectangle =2(a+b)
i.e. 74= 2(a+b)
i.e. 2(a+b) =74
a+b= 74/2
a+b= 37--------(2

since
\[(a-b)^2 =(a+b)^2 -4ab\]
\[(a-b)^2 =(37)^2 -4 \times 102=1369-408=961\]
\[(a-b)^2 =961 \rightarrow a-b = \sqrt{961}\]
\[a-b =31---------(3\]

Adding (2 and (3 we find

a+b= 37
a-b =3
-------------
2a=40
a=20
from eq (2 we find
20+b= 37
b= 37-20
b= 17
Thus the dimensions i.e. length and width of the rectangle are 20 and 17 units respectively.

Answer 2

@nancycruz

Answer 3

+ve solution of a2-37a+102=0