Question

7.4 assignment
possible points: 0.71
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 =

Explanation:

Step1: Identify Proportional Sides

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: \(PQ = 4\), \(QR = 5\), \(XY = 4(n + 1)\), \(YZ = 7n - 1\).
We get \(\frac{4}{4(n + 1)}=\frac{5}{7n - 1}\).

Step2: Cross - Multiply

Cross - multiplying the proportion \(\frac{4}{4(n + 1)}=\frac{5}{7n - 1}\) gives:
\(4(7n - 1)=5\times4(n + 1)\)

Step3: Simplify Both Sides

Simplify the left - hand side: \(4(7n - 1)=28n-4\)
Simplify the right - hand side: \(5\times4(n + 1)=20(n + 1)=20n + 20\)
So, the equation becomes \(28n-4 = 20n+20\)

Step4: Solve for n

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

Answer:

\(3\)