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