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9. if δxyz ~ δrst, find rs. triangle xyz with side xy labeled 5x - 3, s…

Question

  1. if δxyz ~ δrst, find rs.

triangle xyz with side xy labeled 5x - 3, side xz labeled 60. triangle rst with side rt labeled 40, side rs labeled 3x + 2. handwritten equations: (5x - 3)/(3x + 2) = 60/40, 40(5x - 3) = 60(3x + 2), (incomplete step) 19? = 18?

Explanation:

Step1: Set up proportion for similar triangles

Since \(\triangle XYZ \sim \triangle RST\), the corresponding sides are proportional. So \(\frac{XY}{RS}=\frac{XZ}{RT}\), which gives \(\frac{5x - 3}{3x + 2}=\frac{60}{40}\).

Step2: Simplify the proportion

Simplify \(\frac{60}{40}\) to \(\frac{3}{2}\), so the equation becomes \(\frac{5x - 3}{3x + 2}=\frac{3}{2}\).

Step3: Cross - multiply

Cross - multiplying gives \(2(5x - 3)=3(3x + 2)\).

Step4: Expand both sides

Expanding, we get \(10x-6 = 9x + 6\).

Step5: Solve for x

Subtract \(9x\) from both sides: \(10x-9x-6=9x - 9x+6\), which simplifies to \(x - 6=6\). Then add 6 to both sides: \(x=6 + 6=12\).

Step6: Find the length of RS

Now, substitute \(x = 12\) into the expression for \(RS\), which is \(3x + 2\). So \(RS=3\times12+2=36 + 2=38\).

Answer:

\(RS = 38\)