Question

algebra find the value of n that makes $\triangle pqr \sim \triangle xyz$ when $pq = 4$, $qr = 5$, $xy = 4(n + 1)$, $yz = 7n - 1$, and $\angle q \cong \angle y$. (hint: draw a picture)
$n = \square$

Explanation:

Step1: Identify Similar Triangles Proportion

Since \(\triangle PQR \sim \triangle XYZ\) and \(\angle Q \cong \angle Y\), the sides adjacent to these angles are proportional. So, \(\frac{PQ}{XY}=\frac{QR}{YZ}\).
Substitute the given values: \(\frac{4}{4(n + 1)}=\frac{5}{7n - 1}\).

Step2: Cross - Multiply to Solve for n

Cross - multiply the proportion: \(4(7n - 1)=5\times4(n + 1)\).
Simplify both sides: \(28n-4 = 20(n + 1)\).
Expand the right side: \(28n-4=20n + 20\).

Step3: Isolate the Variable n

Subtract \(20n\) from both sides: \(28n-20n-4=20n-20n + 20\), which gives \(8n-4 = 20\).
Add 4 to both sides: \(8n-4 + 4=20 + 4\), so \(8n=24\).
Divide both sides by 8: \(n=\frac{24}{8}=3\).

Answer:

\(3\)