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if $\\angle qrt \\cong \\angle srt$, $qt = 8c - 40$, and $st = c + 37$,…

Question

if $\angle qrt \cong \angle srt$, $qt = 8c - 40$, and $st = c + 37$, what is the value of c

Explanation:

Step1: Identify the theorem

Since \(\angle QRT \cong \angle SRT\), and \(RT\) is common, also \(\angle RST = \angle RQT = 90^\circ\) (from the right angles in the diagram), by the Angle - Angle - Side (AAS) congruence criterion, \(\triangle RST\cong\triangle RQT\). So, \(QT = ST\) (corresponding parts of congruent triangles are equal).

Step2: Set up the equation

Given \(QT = 8c - 40\) and \(ST = c + 37\), and \(QT = ST\), we set up the equation:
\(8c-40=c + 37\)

Step3: Solve for \(c\)

Subtract \(c\) from both sides:
\(8c - c-40=c - c+ 37\)
\(7c-40 = 37\)

Add 40 to both sides:
\(7c-40 + 40=37 + 40\)
\(7c=77\)

Divide both sides by 7:
\(c=\frac{77}{7}=11\)

Answer:

\(11\)