UKMT Senior Challenge-VII

Let $Q$ represent the measure of $\angle PQR$ (in degrees). The measure of $\angle PSR$ is $180-Q$ (I'll leave that to you to prove). Let $a$ be the side length of the rhombus. Applying the law of cosines to triangles $\triangle PQR$ and $\triangle PSR$ , I get the equations: $d_1=\sqrt{a^2+a^2-2a^2\cos{(Q)}}$ and $d_2=\sqrt{a^2+a^2-2a^2\cos{(180-Q)}}$ . Simplifying... $d_1=a\sqrt{2(1-\cos{Q})}$ and $d_2=a\sqrt{2(1+\cos{Q})}$ . Now, applying the condition that the length of the one of the sides of the rhombus is equal to the geometric mean of the lengths of the diagonals, I get $a=\sqrt{a\sqrt{2(1-\cos{(Q)})}a\sqrt{2(1+\cos{(Q)})}} \Rightarrow a^2=2a^2\sqrt{1-\cos^2{(Q)}}\Rightarrow \frac{1}{2}=\sqrt{\sin^2{(Q)}} \Rightarrow \frac{1}{2}=\sin{(Q)}$ . Since $Q$ is obtuse, $Q=150^\circ$ .

Quite easy one...
We know that area of triangle is 1/2absin(C)
Here, area is a^2/2 (by using the gm property given in the question)
Also area is a^2{sin(S)} getting sin (S)=1/2 and S=150