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Home < Algebra Starter Lessons < 8 Unifying Theme For Algebra << 4 Rectangle Area and Like Formulas Backwards
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Forward and Backward Use of Rectangle Area Formula A=WLForward UseExample 1. Find the area of a 5.5 m by 6 m rectangle. A Solution which shows work step by step follows. Remember to copy the format. Solution Draw a Rectangle.
Data identification step. The rectangle has width $W = 5.5 \mbox{ m}$ and Length $L = 6\mbox{ m}$. This step could be done on the diagram.
Formula Evaluation Step. Rectangle area \begin{eqnarray*} A &=& W \times L \\ &=& (5.5 \mbox{ m})\times (6 \mbox{ m}) \\ &=& (5.5 \times 6 ) \mbox{ m}^2 \\ &=& 33 \mbox{ m}^2 \\ \\ &=& 33 \mbox{ square meters} \end{eqnarray*}
Backward Use - NumericallyExample 2. The area of a seven unit long rectangle is 35 square units. Find its width. Arithmetic Solution Draw a Rectangle.
The rectangle area calculation formula \[ A = W L \]Substitution of the given values in this formula yields \[ 35 \mbox{ units}^2 = W \times 7 \mbox{ units} \]Here we have a simple equation in W. The latter equation is equivalent to \[ \frac {35 \mbox{ units}^2}{7 \mbox{ units}} = \frac{ W \times 7 \mbox{ units}} {7 \mbox{ units}} \]By simplication, the latter equation in turn is equivalent to \[ 5 \mbox{ units} = W \]and thus to \[ W = 5 \mbox{ units} \]It was possible to do the calculations without carrying the units through them, But in science and in business calculations, keeping the units with or close to the calculations is an algebraic way aid to avoid errrors in unit handling. Backward Use - AlgebraicallyExample 2 - Generalization. When the length and area of a rectangle are given or known, how can one find its width. Arithmetic Solution Draw a Rectangle.
The rectangle area formula says area \[ A = W L \]In it, length L and area A are supposedly known or given. Swap sides to get the equivalent formula \[ W L = A \]with unknown W on the left hand side. This step is optional. We could have started with latter equation instead of $A=WL$ To isolate W, that is find a formula for it, multiply both sides by $\frac1L$. That is equivalent to dividing both sides by L. The result is \[ \frac {W L}{L} = \frac{A}L \]Keeping the right hand side as is, simplification of the left hand side give \[ W = \frac{A}L \]For a numerical example, let us revisit the numerical example above. It it, A = 35 square units and L = 7 units. The formula for W gives \begin{eqnarray*} W & =& \frac{A}L \\ &=& \frac {35 \mbox{ units}^2}{7 \mbox{ units}} \\ & =& 5 \mbox{ units} \end{eqnarray*} like before. But there is a difference. The formula for W, one can be used in multiple cases where A and L are known or given before W. In my school days, I remember knowing how to derive a formula for W - or some other unknown- while sitting in classes where the instructor gave multiple exercises in which one was expected to plug in numbers, and solve for W numerically. That struck me as a great waste of time, doubly so because I did not know then how to show my instructor nor fellow students how to derive the formula for W - or some other quantity. Once I had a formula for W, and knew how to derive it, the plug-in exercises were not my liking. ,p> Here and in the following lessons on the forward and backward use of formulas, the numerical exercises are intended to provide an arithmetic pattern to follow, and to describe or recast algebraically. Here many similar examples follow. Sooner or later, the hope is that you that the algebraic approach will become natural for you the reader. That will introduce the power of algebra to solve many problems of the same form at once. After that, the limitations of this power will have be seen. The Equation A = BC forwards and backwardsThe physical relations \begin{eqnarray*} \mbox{rectangle area} &=& \mbox{width} \times \mbox{width} \\ \mbox{distance} &=& \mbox{average speed} \times \mbox{travel time} \\ \mbox{voltage} &=& \mbox{Resistance} \times \mbox{Current} \end{eqnarray*} may be known to your or not from geometry and from physics. These relations all have the same algebraic form \[ A = B \times C\]where B and C may be two lengths, speed and time, or resistance and current. Accept the form. Do not worry if you have not yet seen all three examples of the form. The direct use of the form \[ A = B \times C\]would be to calculate A from values of B and C when the latter are given or known. One indirect use would be to find the values of B given the values of A and C. Another indirect use use would be to find the values of B given the values of A and C. We may derive formulas for that. The assumption \[ A = B \times C\]implies \begin{eqnarray*} \frac AC &=\frac {B \times C} C &= B \\ \frac AB &=\frac {B \times C} B &= C \end{eqnarray*} Read backwards, we see that \begin{eqnarray*} B &=& \frac AC \\ C &=& \frac AB \end{eqnarray*} Try to understand the algebraic reasoning given above. The aim here is to develop by example the ability to describe calculations with letters and symbols instead of or besides doing arithmetic. More examples will follow. Changing the SubjectIn North America, obtaining a formula for C given a formula $A = BC$ is called obtaining a literal solution. In the United Kingdom, finding a formula for C given $A = BC$ is called changing the subject of the equation. In the equation, $A = BC$, the number or quantity A is the subject. Implying that $C = \frac AB$ provides a new equation in which C is the subject. |
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Home < Algebra Starter Lessons < 8 Unifying Theme For Algebra << 4 Rectangle Area and Like Formulas Backwards
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