# A piece of wire x cm long is to be cut into two pieces, each to bent to be a square. the length...

## Question:

A piece of wire x cm long is to be cut into two pieces, each to bent to be a square. the length of a side of one square is to be 9 times the length of a side of the other . Express the sum of the areas of two squares in term of x .

## Algebraic Expressions:

An algebraic expression is a function that relates two variables in terms of each other. This implies that as one varies the other will also change subject to the functional relation between them. An algebraic expression will determine the value of one variable in terms of another so that we can evaluate one for any given value of the other.

We can form algebraic expressions from real situations described in words, which can be used in mathematical computations to solve problems.

Suppose that the pieces of wire are of lengths {eq}y, \;x - y {/eq}, respectively. Then the above lengths are also the value of the perimeter of the two squares, respectively.

The lengths of the sides of the two squares would be, {eq}\displaystyle \frac{y}{4} {/eq} and {eq}\displaystyle \frac{x - y}{4} {/eq}, since the four sides of a square are equal.

We have two variables, x and y. We are given the following condition to try to find y in terms of x, i.e.: the length of a side of one square is to be 9 times the length of a side of the other.

Suppose that the square with perimeter y has the longer sides.

Expressing the above information algebraically, we get.

{eq}\displaystyle \frac{y}{4} = 9 \bigg( \frac{x - y}{4} \bigg) {/eq}

The sum of the areas of the two squares should be expressed in terms of the variable x.

So we must solve for the variable y that we introduced, using the above relation:

{eq}\begin{align*} \frac{y}{4} &= 9 \bigg(\frac{x - y}{4} \bigg) \\ \frac{y}{4} &= \frac{9x}{4} - \frac{9y}{4}\\ \frac{y}{4} + \frac{9y}{4} &= \frac{9x}{4} \\ \frac{10y}{4} &= \frac{9x}{4} \\ y &= \frac{9x}{4} \times \frac{4}{10} \\ y &= \frac{9x}{10} \\ \end{align*} {/eq}

Therefore the length of sides of the squares are as follows:

{eq}\begin{align*} s_1 &= \frac{y}{4}\\ s_1 &= \frac{9x}{40} \\ s_2 &= \frac{x - y}{4} \\ s_2 &= \frac{x - \frac{9x}{10} }{4} \\ s_2 &= \frac{ \frac{10 x - 9x}{10} }{4} \\ s_2 &= \frac{x }{40} \\ \end{align*} {/eq}

Therefore the sum of the areas of both the squares would be,

{eq}\begin{align*} A &= s_1^2 + s_2^2 \\ &= \bigg( \frac{9x}{40}\bigg)^2 + \bigg(\frac{x }{40}\bigg)^2 \\ &= \frac{81x^2}{40^2} + \frac{x^2}{40^2} \\ &= \frac{82x^2}{40^2} \\ A(x) &= \frac{41x^2}{800} \\ \end{align*} {/eq} 