Question

Two equal circles of radius 5 cm are touching each other at point \(P\). A common tangent to the circles touches them at points \(Q\) and \(R\). A square is drawn inside triangle \(PQR\) such that one of its sides is resting on this common tangent and the other two vertices touch the two circles. Find the area of the square.

Hint:

When figures involve tangency, always apply the distance formula using the circle’s center and radius.

Answer 1

Correct Answer: 4

Step 1: Place the circles geometrically.
Let the radius of each circle be \(5\) cm.
The common tangent touches the circles at points \(Q\) and \(R\).
The centers of the circles are at \((-5,5)\) and \((5,5)\).
The circles touch each other at point \(P(0,5)\).
Step 2: Assume side of square.
Let the side of the square be \(s\).
The square rests on the tangent, so its bottom side lies on the line \(y = 0\).
The top vertices of the square are at \(\left(\pm \frac{s}{2}, s\right)\).
Step 3: Use distance formula for tangency.
The top left vertex touches the left circle.
Distance from center \((-5,5)\) to \(\left(-\frac{s}{2}, s\right)\) equals radius \(5\).
\[ \left(5 - \frac{s}{2}\right)^2 + (s - 5)^2 = 25 \]
Step 4: Solve the equation.
\[ \left(5 - \frac{s}{2}\right)^2 + (s - 5)^2 = 25 \] \[ \Rightarrow s^2 - 12s + 20 = 0 \] \[ \Rightarrow s = 2 \text{ cm or } 10 \text{ cm} \]
Step 5: Choose the valid solution.
Since the square must lie inside triangle \(PQR\), only \(s = 2\) cm is possible.
Final Answer:
\[ \boxed{4 \text{ cm}^2} \]