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If πΉπ½ is equal to six π₯ plus eight π¦, πΉπ is equal to three π₯ plus five π¦, πΊπ» equals 42, and πΊπ equals 24, what values of π₯ and π¦ make parallelogram πΉπΊπ»π½ a rectangle?
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So weβve been given a lot of information in the question about different lengths within the diagram.
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So letβs begin by transferring this information onto the diagram itself.
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Weβre told that πΉπ½ is equal to six π₯ plus eight π¦.
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And the opposite side, πΊπ», is equal to 42.
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Weβre also given some information about parts of the diagonals of this parallelogram, which is that πΉπ is equal to three π₯ plus five π¦ and πΊπ is equal to 24.
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Now weβre asked what values of π₯ and π¦ make this parallelogram into a rectangle.
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So we need to consider what properties a rectangle has that arenβt true of parallelograms in general.
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Now there are a couple of properties.
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But seeing as weβve been given the lengths of part of the diagonals of this parallelogram, the ones that are relevant here are these: firstly, that the diagonals of a rectangle are equal in length; secondly, they bisect each other, which means they cut each other in half.
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This means then that all four of the line segments going from a vertex of the rectangle to the centre point π are equal in length.
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Specifically taking the two for which weβve been given expressions or values, we can write down the equation three π₯ plus five π¦ is equal to 24.
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And so we have an equation connecting the values of π₯ and π¦.
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However, this is only one equation with two unknowns.
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And so this isnβt sufficient to be able to solve it.
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Letβs think about what other information weβve been given.
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Remember in any parallelogram, the lengths of opposite sides are equal.
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Therefore, the sides πΉπ½ and πΊπ» are equal in length.
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So we can write down another equation: six π₯ plus eight π¦ is equal to 42.
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Now we have a pair of simultaneous equations in π₯ and π¦, which we can solve in order to determine the values that make the first equation true and therefore make the parallelogram a rectangle.
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In this second equation, all of the coefficients are even numbers, which means the whole equation can be divided by two, giving the simpler equation three π₯ plus four π¦ is equal to 21.
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Weβll now look at solving this simpler equation, which Iβve labelled equation two, along with the first equation, equation one.
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We first noticed that both equations have the same coefficient of π₯, itβs three, which means that if I subtract equation two from equation one, this will eliminate the π₯ terms.
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So if I go ahead and do this, then as intended the π₯ terms will cancel out.
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Three π₯ minus three π₯ give no π₯s.
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Five π¦ minus four π¦ gives one π¦, which we just write as π¦.
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And 24 minus 21 gives three.
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So we now have the equation π¦ is equal to three.
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And in fact we found the value of π¦.
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In order to find the value of π₯, we can substitute this value of π¦ into any of the equations.
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Iβm going to choose to substitute into the equation labelled one.
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Making this substitution gives three π₯ plus five multiplied by three is equal to 24.
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Five times three is 15.
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So we have three π₯ plus 15 is equal to 24.
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Next, I subtract 15 from each side of the equation, giving three π₯ is equal to nine.
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The final step is to divide both sides of the equation by three, giving π₯ is equal to three.
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So we found the values of π₯ and π¦ that make the parallelogram πΉπΊπ»π½ a rectangle.
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Theyβre both equal to three.
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Remember, the key facts we used in this question, which are properties of rectangles but not all parallelograms, are these: the diagonals of a rectangle equal in length and they bisect each other.