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Given the triangle πΉπΊπ» is similar to triangle πππ, if six units are added to the length of each side, are the new triangles similar?
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In this question, weβre told that two triangles are similar.
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And we can recall that similar triangles, like any similar polygons, will have corresponding pairs of angles equal and corresponding pairs of sides in proportion.
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If we look at one of the sides π»πΊ, it will be corresponding with the length ππ in the larger triangle.
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And we can see that the lengths here are π on the smaller triangle and three π on the larger triangle.
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The side πΉπ» corresponds to the side ππ, and the lengths are π and three π, respectively.
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The final pair of corresponding sides are πΉπΊ and ππ, and these sides are π and three π.
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If we wanted to find the scale factor of the smaller triangle πΉπΊπ» to the larger triangle πππ, we can see that all the lengths on the smaller triangle must be multiplied by three.
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We now need to consider if six units are added to the length of each side of both triangles, will the two new triangles be similar?
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So, letβs say that instead of having π, π, and π on the smaller triangle, we have π plus six, π plus six, and π plus six.
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On the larger triangle, weβd have three π plus six, three π plus six, and three π plus six.
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At this point, we might think that we do have similar triangles.
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After all, all of the proportions would be the same, except theyβd just have an extra six units added on.
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But letβs consider if we take one of these lengths, say the length π plus six, and actually multiply it by three.
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When we multiply π plus six by three, we must remember that itβs the whole of π plus six thatβs multiplied by three.
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In order to distribute this three across the parentheses, weβll have three π plus three times six, which is 18.
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And on the larger triangle, we have the length of three π plus six and not three π plus 18.
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The same reasoning can be applied to the other two sides.
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For example, ππ would need to be three π plus 18 and ππ would need to be three π plus 18.
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If we return to the definition of similar shapes, we donβt have corresponding sides in proportion, so these two triangles would not be similar.
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We could therefore give the answer to this question as no.
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If you donβt like the use of algebra so much, there is another approach to answering this question.
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Letβs take some actual numerical values for π, π, and π.
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As this is a right triangle, we could use the Pythagorean triple three, four, five.
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The length of ππ would be three times three, which is nine.
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ππ would be three times four, which is 12.
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And ππ would be three times five, which is 15.
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If we then added six to all the lengths on the smaller triangle, weβd have the lengths of nine, 10, and 11, which does actually mean weβve lost this Pythagorean triple, but letβs see what happens anyway.
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On the larger triangle, when we add six units, weβd have the lengths of 15, 18, and 21.
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To check if we have corresponding sides in the same proportion, weβd need to check if 15 over nine is equal to 18 over 10 is equal to 21 over 11.
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And no, these do not have the same proportion.
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So, weβve confirmed that the answer would be no; these two new triangles would not be similar.