Question

1. if \\( \triangle p q r \sim \triangle s q t \\), find the value of \\( x \\).

Explanation:

Step1: Identify corresponding sides

Since \(\triangle PQR \sim \triangle SQT\), the ratios of corresponding sides are equal. So, \(\frac{ST}{PR}=\frac{QS}{QP}\).
\(ST = x - 9\), \(PR = 21\), \(QS = 8\), \(QP = 8 + x + 5=x + 13\). So we get the proportion \(\frac{x - 9}{21}=\frac{8}{x + 13}\).

Step2: Cross - multiply

Cross - multiply the proportion \(\frac{x - 9}{21}=\frac{8}{x + 13}\) to get \((x - 9)(x + 13)=21\times8\).

Step3: Expand and simplify

Expand the left - hand side: \(x^{2}+13x-9x - 117 = 168\).
Simplify the left - hand side: \(x^{2}+4x-117 - 168 = 0\).
So, \(x^{2}+4x-285 = 0\).

Step4: Solve the quadratic equation

Factor the quadratic equation \(x^{2}+4x - 285=0\). We need to find two numbers that multiply to \(- 285\) and add up to \(4\). The numbers are \(19\) and \(-15\) since \(19\times(-15)=-285\) and \(19+( - 15)=4\).
So, \(x^{2}+19x-15x - 285 = 0\).
Factor by grouping: \(x(x + 19)-15(x + 19)=0\).
Then, \((x + 19)(x - 15)=0\).
Setting each factor equal to zero gives \(x=-19\) or \(x = 15\). But since the length cannot be negative (as \(x-9\) and \(x + 13\) represent lengths), we discard \(x=-19\).

Answer:

\(x = 15\)