03 Point P Inside an Isosceles Right Triangle

$18x \cos \theta = 9^2 + x^2 - 15^2$

$\cos \theta = \dfrac{x^2 - 144}{18x}$
 

Apply cosine Law to ΔBPC
$12^2 = 9^2 + x^2 - 2(9x) \cos (90^\circ - \theta)$

$18x \cos (90^\circ - \theta) = 9^2 + x^2 - 12^2$

$\cos (90^\circ - \theta) = \dfrac{x^2 - 63}{18x}$

$\sin \theta = \dfrac{x^2 - 63}{18x}$
 

$\sin^2 \theta + \cos^2 \theta = 1$

$\left( \dfrac{x^2 - 63}{18x} \right)^2 + \left( \dfrac{x^2 - 144}{18x} \right)^2 = 1$

$\dfrac{(x^2 - 63)^2}{324x^2} + \dfrac{(x^2 - 144)^2}{324x^2} = 1$

$(x^2 - 63)^2 + (x^2 - 144)^2 = 324x^2$

$(x^4 - 126x^2 + 3969) + (x^4 - 288x^2 + 20736) = 324x^2$

$2x^4 - 738x^2 + 24705 = 0$

$x^2 = 331.77 ~ \text{ and } ~ 37.23$

$x = 18.21 ~ \text{ and } ~ 6.10$
 

Use x = 18.21 cm
$A = \frac{1}{2}x^2 = \frac{1}{2}(18.21^2)$

$A = 165.80 ~ \text{cm}^2$           answer

$\cos \alpha = \dfrac{15^2 + (9\sqrt{2})^2 - 12^2}{2(15)(9\sqrt{2})}$

$\alpha = 50.476^\circ$
 

Apply Cosine Law to ΔBDC
$x^2 = 9^2 + 15^2 - 2(9)(15) \cos (45^\circ + \alpha)$

$x^2 = 9^2 + 15^2 - 2(9)(15) \cos 95.476^\circ$

$x = 18.21 ~ \text{cm}$
 

$A = \frac{1}{2}x^2 = \frac{1}{2}(18.21^2)$

$A = 165.80 ~ \text{cm}^2$           answer