Find the length and width of a rectangle whose perimeter is 40...

Question

Find the length and width of a rectangle whose perimeter is 40 feet and whose area is 96 square feet.

Accepted Answer

Welcome back. I am so glad you're here. We are told about this rectangular wall And we're told that it has a perimeter of 50 ft in an area of 100 square feet and we're to sulfur its dimensions or its length and its width. So how did we figure out what the perimeter is? Well, the perimeter is taking the length plus the width, plus the length, plus the width. Or adding up all four sides. Or in other words it's two times the length plus two times the width. And for the area while we take the length times the width. And now we've got a system of equations and we can use the parts with the number equal to our variables, the length and the width to solve for the length and the width. I'm going to solve this using substitution. I'm going to rewrite that first equation. So two L plus two W equals 50. I'm going to solve for W. Or excuse me solve for L. By subtracting two W from both sides. So that would be to L equals 50 minus two W. Now dividing both sides by two, solving for W. Give for L. Excuse me solving for L gives us minus W. And now where I have an L. In the second equation, I can substitute this 25 minus W. So now this would read 100 equals not L but 25 minus W. Times W. I can distribute the W on the right hand side of the equation. And now this would read 25 W minus W squared and I'd like to get this W squared to be positive. So I'm going to add that to both sides and I'd like the right hand side equal to zero. So I'm going to subtract 25 W from both sides. So now this will read starting with the w squared w squared minus 25 W plus 100 equals zero. And I can factor this. So what times what gives us W square this W times W and what two numbers when multiplied. Give us positive 100 but add up to a negative 25 that's going to be a minus 20 and a minus five. And now I've got two factors and I can set both of them equal to zero. So w minus 20 equals zero and w minus five equals zero. And I can solve for the width for the first one adding 20 to both sides, and the second one adding five to both sides. So the first one I solve for a width of 20 there's one solution For the second one, I get a width of five. So I've got two different widths and I can set up a solution set and find out what their corresponding lengths would be. So if the width is 20, what would the length b. And if the width is five, what would the length b? I can use either of the two original equations, but I'm going to use the 100. The 2nd 1 100 equals L times W. To solve For the length. And so for the first one substituting 20 in for the width and the second one substituting in five for the width. The first one will read 100 equals length times 20. I can divide both sides by 20 to get the length by itself. 100 divided by 20 equals five. So when the width is 20 the length is five. Now for the other one I get 100 equals all times not W. But now the W is going to be five and dividing both sides by five to solve four L. Gives me 100 divided by five is 20. So when the width is five the length is and I can look at my answer choices and this corresponds with answer choice B. Well done, we'll catch on the next one.

Find the length and width of a rectangle whose perimeter is 40 feet and whose area is 96 square feet.

Key Concepts

Perimeter of a Rectangle

Area of a Rectangle

Solving Systems of Equations

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