Question

what values of a and b make $\triangle rst \cong \triangle hig$?

(image of two right triangles: $\triangle rst$ with right angle at t, side st labeled $3a - 17$, side sr labeled $3a + b + 14$; $\triangle hig$ with right angle at g, side ig labeled $2a$, side hi labeled $a + 2b + 34$)

$a = \square$
$b = \square$

Explanation:

Step1: Identify corresponding sides

Since \(\triangle RST \cong \triangle HIG\), their corresponding sides are equal. The right - angled sides of \(\triangle RST\) are \(ST = 3a - 17\) and \(RT=3a + b+14\), and the right - angled sides of \(\triangle HIG\) are \(IG = 2a\) and \(HG=a + 2b+34\). The right - angled sides of congruent right - triangles are corresponding, so we can set up the following system of equations:
\(

$$\begin{cases}3a-17 = 2a\\3a + b+14=a + 2b+34\end{cases}$$

\)

Step2: Solve for \(a\)

From the first equation \(3a-17 = 2a\), subtract \(2a\) from both sides:
\(3a-2a-17=2a - 2a\)
\(a-17 = 0\)
Add 17 to both sides:
\(a=17\)

Step3: Substitute \(a = 17\) into the second equation and solve for \(b\)

Substitute \(a = 17\) into \(3a + b+14=a + 2b+34\):
\(3\times17 + b+14=17 + 2b+34\)
First, calculate the left - hand side: \(51 + b+14=b + 65\)
Then, calculate the right - hand side: \(17+34 + 2b=2b + 51\)
So we have the equation \(b + 65=2b + 51\)
Subtract \(b\) from both sides:
\(b - b+65=2b - b+51\)
\(65=b + 51\)
Subtract 51 from both sides:
\(b=65 - 51=14\)

Answer:

\(a = 17\), \(b = 14\)