### Video Transcript

The area of the shaded region in the figure below is three π₯ squared π¦ squared plus 10π₯π¦ squared centimeters. By considering the areas of the rectangles π΄π΅πΆπ· and ππΈππΉ, find the length of line segment πΉπ.

Letβs begin by highlighting the line segment πΉπ on the diagram. The question gives us an expression that describes the area of the shaded region. So thatβs the blue region on our diagram. And it tells us to use this alongside the areas of the two rectangles π΄π΅πΆπ· and ππΈππΉ to find the line segment πΉπ. 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 ππΈππΉ from the area of the outer rectangle π΄π΅πΆπ·.

And so, letβs begin by calculating the area of the larger rectangle π΄π΅πΆπ·. The area of a rectangle is base times height or length times width. One of these dimensions is given to us. Itβs simply six π₯π¦. Weβre 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 π·πΉ which is eight centimeters. πΉπ is π₯π¦, and ππΆ is eight. And so, this dimension in our rectangle will be eight plus π₯π¦ plus eight.

And therefore, the area is six π₯π¦ multiplied by eight plus π₯π¦ plus eight. Eight plus π₯π¦ plus eight can be written as 16 plus π₯π¦. And, of course, because a multiplication symbol looks a lot like an π₯, we tend not to include them in algebra. And so, we write this as six π₯π¦ times 16 plus π₯π¦.

Letβs now consider the dimensions of our smaller rectangle; itβs ππΈππΉ. We can see the width of this rectangle is π₯π¦ centimeters. The other dimension of this rectangle is the length weβre trying to find. Itβs the length of line segment πΉπ. So, letβs define that as being equal to π§, or π§ centimeters. And this means the area of this rectangle must be π₯π¦ times π§ or π₯π¦π§. We said that the shaded region is given by the difference between these two areas. So, itβs six π₯π¦ times 16 plus π₯π¦ minus π₯π¦π§.

And at this point, if we spot that thereβs a common factor, we can factor by π₯π¦. And so, this shaded region is π₯π¦ times 16 times 16 plus π₯π¦ minus π§. We can then distribute the inner pair of parentheses, and we get 96 plus six π₯π¦. But of course, the question tells us that the shaded region has an area of three π₯ squared π¦ squared plus 10π₯π¦. So, we can say that these two expressions must be equivalent.

Now, weβre looking to find the value of π§. And so, weβre going to try and rearrange to make π§ the subject. We clear some space, and then we begin by dividing both sides of this equation by π₯π¦. Now, of course, when we do that to the right-hand side, weβre just left with 96 plus six π₯π¦ minus π§. Now, there are a number of ways we can divide by π₯π¦ on that left-hand side. Weβre going to use the bus stop method. The dividend, thatβs three π₯ squared π¦ squared plus 10π₯π¦, goes inside the bus stop. And the divisor goes on the outside. And then, we divide term by term.

Letβs consider π₯π¦ as being equal to one π₯π¦. Then, when we divide three by one, we get three. π₯ squared divided by π₯ is π₯. And π¦ squared divided by π¦ is π¦. Similarly, 10 divided by one is 10, π₯ divided by π₯ is one, and π¦ divided by π¦ is also one. So, this division leaves us with three π₯π¦ plus 10. And now, our equation is equal to three π₯π¦ plus 10 equals 96 plus six π₯π¦ minus π§.

Note that weβre usually really careful when we divide by a variable. If we divide by a variable, we need to be absolutely certain it canβt be equal to zero, since dividing by zero gives us a result thatβs undefined. However, weβve been given π₯π¦ as being a dimension. And this isnβt equal to zero, as we can see. And so, weβre absolutely fine to divide through by this value.

Letβs make π§ the subject by adding it to both sides. When we do, we get π§ plus three π₯π¦ plus 10 equals 96 plus six π₯π¦. In our next step, weβre going to subtract both three π₯π¦ and 10 from both sides. That leaves us with π§ equals 86 plus three π₯π¦ or three π₯π¦ plus 86. Our units are centimeters.

So, we can say the length of line segment πΉπ is three π₯π¦ plus 86 centimeters.