GK 2019 Q24

Label the intersection of \(WQ\) and \(XP\) as \(V\) and the midpoint of \(WX\) as \(U\). Let the side-length of the square be 1 unit. The area of triangle \(WXR\) is \(\tfrac{1}{2} \times 1 \times 1\) units² = \(\tfrac{1}{2}\) units². Consider triangle \(WXP\) and triangle \(UXV\). These two triangles have the same angles and hence are similar. Since \(UX\) is half of \(WX\), it follows that \(VU\) is half of \(PW\) and hence has length \(\tfrac{1}{4}\) unit. Therefore the area of triangle \(WXV\) is \(\left(\tfrac{1}{2} \times 1 \times \tfrac{1}{4}\right)\) units² = \(\tfrac{1}{8}\) units². Hence the shaded area is \(\left(\tfrac{1}{2} - \tfrac{1}{8}\right)\) units² = \(\tfrac{3}{8}\) units². Therefore the fraction of the square that is shaded is \(\tfrac{3}{8}\).