Question

ABCD is a rectangle. Points \( P \) and \( Q \) lie on \( AD \) and \( AB \) respectively. If triangles \( PAQ, QBC, \) and \( PCD \) all have the same areas and \( BQ = 2 \), then \( AQ = \) ?

• A. \( 1 + \sqrt{5} \)
• B. \( 1 - \sqrt{5} \)
• C. \( \sqrt{7} \)
• D. \( 2\sqrt{7} \)

Hint:

Use area equality and triangle area formula carefully with coordinates. Symmetry and rectangle structure help.

Answer 1

The Correct Option is A

Let the rectangle be of size \( a \times b \). Let \( PAQ = QBC = PCD = A \) (equal areas).
Since these are triangles and total area of rectangle = \( ab \), the 3 triangle areas sum up to the area:
\[ 3A = ab \Rightarrow A = \frac{ab}{3} \] Now use triangle area formulas for coordinates and side lengths. Assign coordinates: \[ A(0, 0), B(a, 0), C(a, b), D(0, b) \] Let \( Q = (x, 0) \), then \( AQ = x \), \( BQ = a - x = 2 \Rightarrow x = a - 2 \) Use triangle area formula for \( \triangle PAQ = \frac{ab}{3} \), and similarly compute using determinants. Solving the resulting quadratic gives: \[ AQ = 1 + \sqrt{5} \] \[ {1 + \sqrt{5}} \]