Find the length and width of a rectangle whose perimeter is 36...

Question

Find the length and width of a rectangle whose perimeter is 36 feet and whose area is 77 square feet.

Accepted Answer

Welcome back. I am so glad you're here. We are told about this rectangular wall, we are told that it has a perimeter of 40 ft And it has an area of 64 square feet were to determine its dimensions or its length and its width. Well, how do we figure out the perimeter? The perimeter is? Well, if we've got a length and width, we're going to take the length plus the width, plus the length plus the width, adding all four sides together. Or in other words the perimeter is twice the length, plus twice the width. How about the area? How do we solve for that? Well, that is taking the length times the width. Well, we have now got a system of equations. We have two equations and we can use them to solve for the length and the width. Let's just focus on the part that has the number equal to our variables that we need L. And W. And I'm going to solve this using substitution, I'm going to start off by rewriting the first equation. Two L plus two W equals 40 and solving for L. So I'm going to subtract two W from both sides. That would be two. L equals 40 minus two W. And dividing both sides by two. That will be the length equals minus W. And now for that substitution where I have an L. In the second equation, I can substitute 20 minus W. So the second equation will now read 64 equals not L but 20 minus W times W. Now I can distribute the w. On the right hand side. So this will read 64 equals 20 W minus W squared. We have a quadratic and I would like to get the highest degree term, the w squared to be positive. And I would like to get the right hand side equal to zero. So I will add w squared to both sides and subtract 20 W from both sides and will have w squared minus W plus 64 equals zero. And we can factor this what times what gives us W square? That's W. Times W. And what two numbers when multiplied together, give us a positive 64 add up to a negative 20. That's going to be a minus 16 minus four. So now we have two factors, we can set both of them equal to zero, W minus 16 equals zero and w minus four equals zero. To solve for W. For the first one adding 16 to both sides, and the second one adding four to both sides, solving for the first one gives us width of and solving for the second one gives us a width of four. And now we've got a solution set, we've got two different options for our width and now we can figure out what the corresponding options for the length would be. We can solve using either of our two original equations. I'd like to use the second one. So 64 equals solving for L and now it's L Times no longer W but now 16. And solving for W, dividing both sides by 16, divided by 16 is four, so when the width is 16 the length is four. Now let's solve for when the width is four, so this will be 64 equals the length times four, and Not W. and dividing both sides by four, 64, divided by 4, 16. So when the width is for the length is 16, while we look at our answer choices, and this matches with answer choice. A. Oops, well done, we'll catch you on the next one.

Find the length and width of a rectangle whose perimeter is 36 feet and whose area is 77 square feet.

Key Concepts

Perimeter of a Rectangle

Area of a Rectangle

Solving Systems of Equations

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