, 2014

Solution 1

Since $\triangle RPS$ is right-angled at $P$, then by the Pythagorean Theorem, $PR^2 + PS^2 = RS^2$ or $PR^2 +18^2 = 30^2$. This gives $PR^2 = 30^2-18^2 = 900 - 324 = 576$, from which $PR=24$, since $PR>0$. Since $P$, $S$ and $Q$ lie on a straight line and $RP$ is perpendicular to this line, then $RP$ is actually a height for $\triangle QRS$ corresponding to base $SQ$. Thus, the area of $\triangle QRS$ is $\frac{1}{2}(24)(14)=168$. Solution 2 Since $\triangle RPS$ is right-angled at $P$, then by the Pythagorean Theorem, $PR^2 + PS^2 = RS^2$ or $PR^2 +18^2 = 30^2$. This gives $PR^2 = 30^2-18^2 = 900 - 324 = 576$, from which $PR=24$, since $PR>0$. The area of $\triangle QRS$ equals the area of $\triangle RPQ$ minus the area of $\triangle RPS$. Since $\triangle RPQ$ is right-angled at $P$, its area is $\frac{1}{2}(PR)(PQ)=\frac{1}{2}(24)(18+14)=12(32)=384$. Since $\triangle RPS$ is right-angled at $P$, its area is $\frac{1}{2}(PR)(PS)=\frac{1}{2}(24)(18)=12(18)=216$. Therefore, the area of $\triangle QRS$ is $384-216=168$.