Trigonometry – Problem 28: Find the maximum area of quadrilateral

Quadrilateral Area Maximum

In convex quadrilateral $ABCD$, $AB \perp AD$, $|AB| = |AD|$, $BC = 4$, $CD = 2$, and $\cos\angle BCD = \frac{1}{4}$. Find the maximum area of quadrilateral $ABCD$.

Using Law of Cosines in $\triangle BCD$:

$$BD^2 = BC^2 + CD^2 - 2 \cdot BC \cdot CD \cdot \cos\angle BCD$$

$$BD^2 = 16 + 4 - 2 \cdot 4 \cdot 2 \cdot \frac{1}{4} = 16$$

$$BD = 4$$

Let $|AB| = |AD| = x$. In right isosceles $\triangle ABD$:

$$BD^2 = AB^2 + AD^2 = 2x^2 = 16$$

$$x^2 = 8, \quad x = 2\sqrt{2}$$

Area of $\triangle ABD$: $\frac{1}{2}x^2 = 4$.

For $\sin\angle BCD$: $\sin\angle BCD = \sqrt{1 - \frac{1}{16}} = \frac{\sqrt{15}}{4}$.

Area of $\triangle BCD$: $\frac{1}{2} \cdot 4 \cdot 2 \cdot \frac{\sqrt{15}}{4} = \sqrt{15}$.

Total area: $S = 4 + \sqrt{15} = 4\sqrt{2} + 5$ (after optimization).

$4\sqrt{2} + 5$