Question

A farmer has a squared field of area \[{{x}^{2}}-26x+169\] square units. He divided the field into two equal parts. Then find the sides of the field in terms of x.

Answer 1

Hint: We know that if the side of the square is equal to a, then the area of the square is equal to \[{{a}^{2}}\]. Let us consider the square field is in the form of a square CDEF. Let us assume the area of the square filed CDEF is equal to A and the side of the square CDEF is equal to a. Now we will find the side of the given square. From the question, we were given that a farmer has a squared field of area \[{{x}^{2}}-26x+169\] square units. He divided the field into two equal parts. So, it is clear that the square CDEF is divided into equal parts. Let us assume a line AB is drawn which divides the square filed into two parts. So, it is clear that AB divides the side CD and side EF into two equal parts. Let us assume the length of AC in the rectangle ABCD is equal to S. So, here S and a give us the sides of the field in terms of x.

Complete step-by-step answer:
From the question, we were given that a farmer has a squared field of area \[{{x}^{2}}-26x+169\] square units. He divided the field into two equal parts.
Now we have to find the side of the square.
We know that if the side of the square is equal to a, then the area of the square is equal to \[{{a}^{2}}\].
Let us consider the square field is in the form of a square CDEF.
Let us assume the area of the square filed CDEF is equal to A and the side of the square CDEF is equal to a.
\[\begin{align}
  & \Rightarrow A={{x}^{2}}-26x+169 \\
 & \Rightarrow {{a}^{2}}={{x}^{2}}-2(13)(x)+{{\left( 13 \right)}^{2}} \\
\end{align}\]
We know that \[{{\left( x-a \right)}^{2}}={{x}^{2}}-2ax+{{a}^{2}}\].
Now we will apply this formula, then we get
\[\begin{align}
  & \Rightarrow {{a}^{2}}={{\left( x-13 \right)}^{2}} \\
 & \Rightarrow a=x-13....(1) \\
\end{align}\]

It is given that the square CDEF is divided into equal parts. Let us assume a line AB is drawn which divides the square field into two parts.
So, it is clear that AB divides the side CD and side EF into two equal parts.
Let us assume the length of AC in rectangle ABCD is equal to S.
\[\Rightarrow S=\dfrac{x-13}{2}.....(2)\]
So, it is clear that the breadth of the rectangle ABCD is equal to \[x-13\] and the length of the rectangle ABCD is equal to \[\dfrac{x-13}{2}\].
Hence, we can say that the length of the sides of the field are \[x-13\] and \[\dfrac{x-13}{2}\] if the field is divided into two parts.

Note: Students may have a misconception that \[{{\left( x+a \right)}^{2}}={{x}^{2}}-2ax+{{a}^{2}}\]. Then we will get that the side of square EFCD is equal to \[x+13\]. But we know that side of square EFCD is equal to \[x-13\]. This will give us a wrong measure of sides of the field which is divided into two parts. So, this misconception should be avoided. No calculation error should be made by a student. If a small mistake is done, wrong results may be obtained.