Question Video: Using Properties of Similar Polygons to Find an Unknown By Forming a Linear Equation | Nagwa Question Video: Using Properties of Similar Polygons to Find an Unknown By Forming a Linear Equation | Nagwa

Question Video: Using Properties of Similar Polygons to Find an Unknown By Forming a Linear Equation Mathematics • First Year of Secondary School

Given that 𝐴𝐵𝐶𝐷 ∼ 𝑍𝑌𝑋𝐿, find the value of 𝑋.

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Video Transcript

Given that 𝐴𝐵𝐶𝐷 is similar to 𝑍𝑌𝑋𝐿, find the value of 𝑋.

We’ve been given that two polygons, or in fact quadrilaterals, 𝐴𝐵𝐶𝐷 and 𝑍𝑌𝑋𝐿 are similar to one another. We recall that similar polygons have two key properties. Firstly, corresponding angles are congruent. And secondly, corresponding sides are proportional.

We can determine which vertices are corresponding to one another by considering the ordering of the letters in the similarity statement. We were told that 𝐴𝐵𝐶𝐷 is similar to 𝑍𝑌𝑋𝐿, so vertex 𝐴 is corresponding to vertex 𝑍, vertex 𝐵 is corresponding to vertex 𝑌, vertex 𝐶 is corresponding to vertex 𝑋, and vertex 𝐷 is corresponding to vertex 𝐿. This also helps us determine which sides on the two polygons are corresponding. The side connecting the vertices 𝐴 and 𝐵 on the smaller polygon is corresponding to the side connecting the vertices 𝑍 and 𝑌 on the larger. And the side connecting vertices 𝐶 and 𝐷 on the smaller polygon is corresponding to the side connecting vertices 𝑋 and 𝐿 on the larger polygon.

We can then use the fact that corresponding sides in similar polygons are proportional to write an equation. Using the pairs of proportional sides we’ve identified, we have 𝐶𝐷 over 𝑋𝐿 is equal to 𝐴𝐵 over 𝑍𝑌. Equivalently, we could write the reciprocal of this; 𝑋𝐿 over 𝐶𝐷 is equal to 𝑍𝑌 over 𝐴𝐵. We can then substitute the lengths or expressions given in the diagram for each of these sides. We have 15 plus two 𝑋 over 246.2 is equal to 75 over 150. This is why I chose to write the proportional relationship this way round rather than its reciprocal because then the unknown 𝑋 is in the numerator of the fraction.

Now the fraction on the right-hand side can be simplified by dividing both the numerator and denominator by 75 to give one-half. This tells us that lengths on the smaller polygon are half as long as the corresponding lengths on the larger polygon. Or the other way round, lengths on the larger polygon are twice as long as the corresponding lengths on the smaller polygon.

We could then approach the problem from a logical point of view, or we can continue with solving the equation we’ve formed. Multiplying both sides of the equation by 246.2 gives 15 plus two 𝑋 equals 246.2 over two, or 123.1. We want to solve for 𝑋, so the next step is to subtract 15 from each side of the equation, giving two 𝑋 equals 108.1. Finally, we can divide both sides of the equation by two to give 𝑋 equals 54.05.

So by recalling that corresponding sides in similar polygons are proportional, and then forming an equation involving the length of two pairs of corresponding sides, we found that the value of the unknown 𝑋 is 54.05.

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