Question

Three villages \(P, Q,\) and \(R\) are located in such a way that the distances are \(PQ=13\) km, \(QR=14\) km, and \(RP=15\) km (see figure). A straight road joins \(Q\) and \(R\). It is proposed to connect \(P\) to this road \(QR\) by constructing another road. What is the minimum possible length (in km) of this connecting road?

• A. 10.5
• B. 11.0
• C. 12.0
• D. 12.5

Hint:

When asked for the shortest distance from a vertex to the opposite side in a triangle with all side lengths known, use Heron’s formula to find the area and then \(h=\dfrac{2\Delta}{\text{base}}\).

Answer 1

The Correct Option is C

Step 1: Interpret the “minimum length.”
The shortest road from \(P\) to the line \(QR\) is the perpendicular distance (altitude) from \(P\) to \(QR\). So we need the altitude from \(P\) to side \(QR\) of \(\triangle PQR\).

Step 2: Area of \(\triangle PQR\) via Heron’s formula.
Let the sides be \(a=QR=14\), \(b=PR=15\), \(c=PQ=13\).
Semiperimeter: \[ s=\frac{a+b+c}{2}=\frac{14+15+13}{2}=21. \] Area: \[ \Delta=\sqrt{s(s-a)(s-b)(s-c)} =\sqrt{\,21\cdot 7\cdot 6\cdot 8\,} =\sqrt{7056}=84. \]

Step 3: Compute the altitude from \(P\) to \(QR\).
If \(h\) is the altitude to side \(QR\), then \[ \Delta=\frac{1}{2}\cdot QR\cdot h \ \Rightarrow\ h=\frac{2\Delta}{QR}=\frac{2\cdot 84}{14}=12\ \text{km}. \] Final Answer:
\[ \boxed{12.0\ \text{km}} \]