Question Video: Using Properties of Similar Triangles to Solve Algebraic Problems | Nagwa Question Video: Using Properties of Similar Triangles to Solve Algebraic Problems | Nagwa

Question Video: Using Properties of Similar Triangles to Solve Algebraic Problems Mathematics • First Year of Secondary School

The given figure shows a triangle 𝐴𝐵𝐶. (1) Work out the value of 𝑥. (2) Work out the value of 𝑦.

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Video Transcript

The given figure shows a triangle 𝐴𝐵𝐶. (1) work out the value of 𝑥. And (2) work out the value of 𝑦.

We’re given a triangle with a line parallel to one side inscribed within it and various lengths of segments of the sides of the triangle. And we’re asked to find the value of 𝑥 and the value of 𝑦.

Let’s begin with part (1), which is finding 𝑥, where we see the two of our side segments involve 𝑥. We note first that a line of length two units inside the triangle is parallel to the side 𝐵𝐶. Now, the side splitter theorem tells us that this line divides the two sides 𝐴𝐶 and 𝐴𝐵 proportionally.

Remember, according to the side splitter theorem, if a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides those sides proportionally. If we label this line segment 𝐷𝐸 in our diagram, we could say that 𝐴𝐷 over 𝐷𝐵 is equal to 𝐴𝐸 over 𝐸𝐶. Substituting in the given lengths, we can form an equation which we can solve for 𝑥. That’s three over two 𝑥 plus three is equal to two over 𝑥 plus five. Now, multiplying both sides by 𝑥 plus five and two 𝑥 plus three and distributing the parentheses, we have three 𝑥 plus 15 equals four 𝑥 plus six. Subtracting three 𝑥 and six from both sides and swapping sides, we then have 𝑥 is equal to nine.

So now, making a note of this and making some space, we can use this value of 𝑥 in part (2) of the question to find the value of 𝑦. Now, since the two triangles created by the intersection of side 𝐷𝐸 share a common angle 𝐴 and the pairs of corresponding angles also created by this line are equal, we can say the triangles 𝐴𝐷𝐸 and 𝐴𝐵𝐶 are similar triangles. In particular, this means that the proportions 𝐴𝐷 over 𝐴𝐵 and 𝐷𝐸 over 𝐵𝐶 are equal. Now, we know that 𝐴𝐵 is equal to the sum 𝐴𝐷 plus 𝐷𝐵. And that’s three plus two 𝑥 plus three. And we have 𝑥 equal to nine from part (1). And so this evaluates to 24.

So now, substituting our values 𝐴𝐷 equals three, 𝐴𝐵 equals 24, 𝐷𝐸 equals two, and 𝐵𝐶 equals 𝑦 into our equation, we have three over 24 is equal to two over 𝑦. Now, multiplying both sides by 𝑦 and by 24 over three, we have 𝑦 equal to 16. Hence, from the given figure, we find that 𝑥 is equal to nine and 𝑦 is equal to 16.

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