# Question Video: Finding the Length of a Side in a Polygon given the Corresponding Side in a Similar Polygon and the Similarity Ratio between Them Mathematics

Given that the two polygons are similar, find the value of π₯.

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### Video Transcript

Given that the two polygons are similar, find the value of π₯.

We know that any similar polygons have corresponding angles that are congruent and corresponding sides that are proportional. Due to the orientation of these shapes, it may not immediately be obvious which sides are corresponding. In order to work this out, it is useful to identify the corresponding angles first. One pair of corresponding sides are π΅πΆ and ππ». A second pair of corresponding sides are therefore πΆπ· and π»π½.

As the corresponding sides are proportional, we know that the ratios two to six and four π₯ minus 37 to two π₯ minus 11 must be equal. Writing this in fractional form, we have two over four π₯ minus 37 is equal to six over two π₯ minus 11. Both of the numerators here are divisible by two. We can then cross multiply to give us one multiplied by two π₯ minus 11 is equal to three multiplied by four π₯ minus 37.

Distributing our parentheses gives us two π₯ minus 11 is equal to 12π₯ minus 111. Adding 111 to both sides of this equation gives us two π₯ plus 100 is equal to 12π₯. We can then subtract two π₯ from both sides of this equation, which gives us 100 is equal to 10π₯. Finally, dividing both sides of this equation by 10 gives us a value of π₯ equal to 10.

We can then substitute this value back into the expressions for the lengths of π΅πΆ and ππ» to check our answer. Four multiplied by 10 is equal to 40. Subtracting 37 from this gives us three. Two multiplied by 10 is equal to 20, and subtracting 11 from this gives us nine. The ratios two to six and three to nine are identical as they can both be simplified to one to three.

An alternative method in this question would be to initially recognize that the scale factor was three. This is because the length π½π» is three times the length of πΆπ·. We could then have set up the equation two π₯ minus 11 is equal to three multiplied by four π₯ minus 37 as the length ππ» is three times the length of π΅πΆ. Following this method would also have got us a value of π₯ equal to 10.