Hello, sallyk57!
Don't feel bad . . . this is not easy.
\(\displaystyle \;\;\)I had to "invent" a solution . . .
Mona and Lisa fields are triangles, neither of them equilateral.
All sides of both triangles measure whole numbers of hectometers.
Two sides of Mona’s field have the same length as two sides of Lisa’s field.
The three angles of Mona’s field are equal to the three angles of Lisa’s field.
The area of Mona’s field is smaller than that of Lisa’s field.
What is the minimum perimeter of the smaller field?
Since the triangles are similar (but not congruent),
\(\displaystyle \;\;\)they must look like this:
Code:
*
* a * * b
c * * a * *
* * * *
*---------------* *-------------------*
b * d
Sides \(\displaystyle a\) and \(\displaystyle b\) are the longer sides of Mona's field.
Sides \(\displaystyle a\) and \(\displaystyle b\) are the shorter sides of Lisa's field.
Let \(\displaystyle c\) be the third side of Mona's field.
Let \(\displaystyle d\) be the third side of Lisa's field.
From the similar triangles, we have: \(\displaystyle \,\frac{c}{a}\,=\,\frac{a}{b}\,=\,\frac{b}{d}\)
From the first two fractions, we have: \(\displaystyle \,a^{^2}\,=\,bc\)
\(\displaystyle \;\;\)That is, \(\displaystyle a\) is the mean proportional of \(\displaystyle b\) and \(\displaystyle c\).
The smallest solution (for nonequal numbers that can form a triangle) is:
\(\displaystyle \;\;\;a\,=\,6,\;b\,=\,4,\;c\,=\,9\)
From the last two fractions, we have: \(\displaystyle \,ad\,=\,b^2\)
\(\displaystyle \;\;\)Plugging in our values, we have: \(\displaystyle 6d\,=\,16\;\;\Rightarrow\;\;d\,=\,\frac{8}{3}\)
But \(\displaystyle d\) is a whole number.
So we will multiply all the dimension by 3: \(\displaystyle \,a\,=\,18,\:b\,=\,12,\:c\,=\,27,\;d\,=\,8\)
These are the smallest triangles that satisfy the problem.
Therefore, the perimeter of Mona's field is: \(\displaystyle \,18\,+\,12\,+\,27\:=\:57\)
