The area of the shaded region in the figure below is three 𝑥 squared 𝑦 squared plus 10𝑥𝑦 squared centimeters. By considering the areas of the rectangles 𝐴𝐵𝐶𝐷 and 𝑀𝐸𝑁𝐹, find the length of line segment 𝐹𝑁.
Let’s begin by highlighting the line segment 𝐹𝑁 on the diagram. The question gives us an expression that describes the area of the shaded region. So that’s the blue region on our diagram. And it tells us to use this alongside the areas of the two rectangles 𝐴𝐵𝐶𝐷 and 𝑀𝐸𝑁𝐹 to find the line segment 𝐹𝑁. Now, there are a whole bunch of ways that we could calculate the area of the shaded region. We could, for example, split it up into three rectangles. Alternatively, we can spot that we can find the area of the shaded region by subtracting the area of the nonshaded rectangle 𝑀𝐸𝑁𝐹 from the area of the outer rectangle 𝐴𝐵𝐶𝐷.
And so, let’s begin by calculating the area of the larger rectangle 𝐴𝐵𝐶𝐷. The area of a rectangle is base times height or length times width. One of these dimensions is given to us. It’s simply six 𝑥𝑦. We’re going to need to, however, calculate the other dimension. It will be given by the sum of the lengths of these three individual line segments. We have line segment 𝐷𝐹 which is eight centimeters. 𝐹𝑀 is 𝑥𝑦, and 𝑀𝐶 is eight. And so, this dimension in our rectangle will be eight plus 𝑥𝑦 plus eight.
And therefore, the area is six 𝑥𝑦 multiplied by eight plus 𝑥𝑦 plus eight. Eight plus 𝑥𝑦 plus eight can be written as 16 plus 𝑥𝑦. And, of course, because a multiplication symbol looks a lot like an 𝑥, we tend not to include them in algebra. And so, we write this as six 𝑥𝑦 times 16 plus 𝑥𝑦.
Let’s now consider the dimensions of our smaller rectangle; it’s 𝑀𝐸𝑁𝐹. We can see the width of this rectangle is 𝑥𝑦 centimeters. The other dimension of this rectangle is the length we’re trying to find. It’s the length of line segment 𝐹𝑁. So, let’s define that as being equal to 𝑧, or 𝑧 centimeters. And this means the area of this rectangle must be 𝑥𝑦 times 𝑧 or 𝑥𝑦𝑧. We said that the shaded region is given by the difference between these two areas. So, it’s six 𝑥𝑦 times 16 plus 𝑥𝑦 minus 𝑥𝑦𝑧.
And at this point, if we spot that there’s a common factor, we can factor by 𝑥𝑦. And so, this shaded region is 𝑥𝑦 times 16 times 16 plus 𝑥𝑦 minus 𝑧. We can then distribute the inner pair of parentheses, and we get 96 plus six 𝑥𝑦. But of course, the question tells us that the shaded region has an area of three 𝑥 squared 𝑦 squared plus 10𝑥𝑦. So, we can say that these two expressions must be equivalent.
Now, we’re looking to find the value of 𝑧. And so, we’re going to try and rearrange to make 𝑧 the subject. We clear some space, and then we begin by dividing both sides of this equation by 𝑥𝑦. Now, of course, when we do that to the right-hand side, we’re just left with 96 plus six 𝑥𝑦 minus 𝑧. Now, there are a number of ways we can divide by 𝑥𝑦 on that left-hand side. We’re going to use the bus stop method. The dividend, that’s three 𝑥 squared 𝑦 squared plus 10𝑥𝑦, goes inside the bus stop. And the divisor goes on the outside. And then, we divide term by term.
Let’s consider 𝑥𝑦 as being equal to one 𝑥𝑦. Then, when we divide three by one, we get three. 𝑥 squared divided by 𝑥 is 𝑥. And 𝑦 squared divided by 𝑦 is 𝑦. Similarly, 10 divided by one is 10, 𝑥 divided by 𝑥 is one, and 𝑦 divided by 𝑦 is also one. So, this division leaves us with three 𝑥𝑦 plus 10. And now, our equation is equal to three 𝑥𝑦 plus 10 equals 96 plus six 𝑥𝑦 minus 𝑧.
Note that we’re usually really careful when we divide by a variable. If we divide by a variable, we need to be absolutely certain it can’t be equal to zero, since dividing by zero gives us a result that’s undefined. However, we’ve been given 𝑥𝑦 as being a dimension. And this isn’t equal to zero, as we can see. And so, we’re absolutely fine to divide through by this value.
Let’s make 𝑧 the subject by adding it to both sides. When we do, we get 𝑧 plus three 𝑥𝑦 plus 10 equals 96 plus six 𝑥𝑦. In our next step, we’re going to subtract both three 𝑥𝑦 and 10 from both sides. That leaves us with 𝑧 equals 86 plus three 𝑥𝑦 or three 𝑥𝑦 plus 86. Our units are centimeters.
So, we can say the length of line segment 𝐹𝑁 is three 𝑥𝑦 plus 86 centimeters.