Question Video: Finding the Perimeter of a Rectangle given Its Area and the Difference between Its Dimensions Mathematics • 9th Grade

What is the perimeter of a rectangle whose length is 7 cm more than its width and whose area is 78 cm²?

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Video Transcript

What is the perimeter of a rectangle whose length is seven centimeters more than its width and whose area is 78 centimeters squared?

Now, in a problem like this, it’s always a good idea to draw our own diagram so that we can visualize the situation. We have a rectangle, so it looks a little something like this. Now, we don’t know its width, so we can use the letter 𝑤 to represent this unknown value. We’re told that the length of the rectangle is seven centimeters more than the width, which means we can express the length in terms of the width. If the width is 𝑤, then the length will be 𝑤 plus seven. So we have expressions for both the length and the width of the rectangle in terms of the same letter, 𝑤. We’re also told that the area of this rectangle is 78 centimeters squared, and we’re asked to find its perimeter.

In order to do this, we’re going to need to know the value of 𝑤. So we can use the information we’ve been given to form an equation. The area of a rectangle is found by multiplying its length by its width. So, using the expressions we have for these two values, we have 𝑤 multiplied by 𝑤 plus seven or 𝑤 plus seven multiplied by 𝑤. But as we know, the area is 78 centimeters squared. We can write this as an equation. We have then 𝑤 multiplied by 𝑤 plus seven equals 78.

The next step is to rearrange this equation slightly, which we’ll do by expanding the brackets or distributing the parentheses as you may call it. 𝑤 multiplied by 𝑤 gives 𝑤 squared, and 𝑤 multiplied by seven gives seven 𝑤. So we have 𝑤 squared plus seven 𝑤 equals 78. The next step is just subtract 78 from each side of the equation, as we want to group all of the terms together on the same side. Doing so gives 𝑤 squared plus seven 𝑤 minus 78 is equal to zero. And now, we see that we have a quadratic equation in 𝑤 in this most easily recognizable form. We need to solve this equation to find the value of 𝑤. So let’s see whether this equation can be factored.

As the coefficient of 𝑤 squared, that’s the number in front of 𝑤 squared, is just one, the first term in each of our factors will just be 𝑤. And to complete our factors, we’re then looking for two numbers whose sum is the coefficient of 𝑤, that’s positive seven, and whose product is the constant term, that’s negative 78. To find these numbers, we can begin by listing all the factor pairs of 78. They are one and 78, two and 39, three and 26, and six and 13. Now, the product needs to be negative 78, which means one of our numbers needs to be negative and the other needs to be positive. And we see that if we use our last factor pair and we choose six to be negative but 13 to be positive. Then we have negative six plus 13 or 13 minus six, which is indeed equal to seven. So the sum of this factor pair would be the correct value.

So we can complete our two sets of parentheses by adding the value negative six to the first and positive 13 to the second. We now have the factored form of our quadratic. And you can confirm that this is indeed correct by redistributing or reexpanding the brackets. The next step in this method is to recall that if the product of two factors is zero, then at least one of those factors must themselves be zero. So we take each factor in turn and set it equal to zero, giving 𝑤 minus six equals zero or 𝑤 plus 13 equals zero. We can then solve the resulting linear equations.

To solve the first equation, we add six to both sides giving 𝑤 equals six. And to solve the second, we subtract 13 from each side giving 𝑤 equals negative 13. So we have two solutions to our quadratic equation. 𝑤 equals six and 𝑤 equals negative 13. However, whilst these are both valid solutions to the quadratic equation, they aren’t both valid in terms of the context of this problem. If we look back at our rectangle, we can see that 𝑤 represents its width. And the width must take a positive value. 𝑤 therefore cannot be equal to negative 13 in this problem. So we eliminate this value. We’ve found then that the value of 𝑤 and the width of our rectangle is six.

Now, finally, we recall that we weren’t just asked to find the width of the rectangle, but we were asked to find its perimeter. So we now know that its width is six centimeters and its length is six plus seven; that’s 13 centimeters. The perimeter is the distance all the way around the edge of this rectangle. So we can either just add up all four sides, 13 plus six plus 13 plus six. Or we could use the formula that the perimeter is equal to twice the sum of the length and the width. So we could have two times six plus 13. In either case, it will, of course, give the same value of 38. The area of this rectangle was given in centimeters squared, so the units for its perimeter will be centimeters. And we’ve found then that the perimeter of the rectangle is 38 centimeters.

To check our answer, we can multiply the length and the width that we’ve calculated together and confirm that the area is indeed equal to 78 centimeters squared.