Question

the figure shows a pair of similar triangles. find the lengths of the sides labeled with the variables, △abc ~ △rst. 2 points. st = . ac = .

Explanation:

Step1: Set up proportion for similar - triangles

Since $\triangle ABC\sim\triangle RST$, the ratios of corresponding sides are equal. So, $\frac{AC}{RT}=\frac{BC}{ST}$. Substituting the given values, we get $\frac{y - 1}{6}=\frac{2}{y}$.

Step2: Cross - multiply

Cross - multiplying the proportion $\frac{y - 1}{6}=\frac{2}{y}$ gives us $(y - 1)y=2\times6$. Expanding the left - hand side, we have $y^{2}-y = 12$.

Step3: Rearrange to quadratic form

Rearranging the equation $y^{2}-y = 12$ to the standard quadratic form $ay^{2}+by + c = 0$, we get $y^{2}-y-12 = 0$.

Step4: Factor the quadratic equation

Factoring $y^{2}-y - 12=0$, we have $(y - 4)(y+3)=0$.

Step5: Solve for y

Setting each factor equal to zero gives $y - 4=0$ or $y + 3=0$. So, $y=4$ or $y=-3$. But since $y$ represents the length of a side of a triangle, $y>0$. So, $y = 4$.

Step6: Find the lengths of the sides

For $ST=y$, so $ST = 4$. For $AC=y - 1$, substituting $y = 4$, we get $AC=4 - 1=3$.

Answer:

$ST = 4$
$AC = 3$