21. The diagram shows a square $P Q R S$ with area $120 \mathrm{~cm}^{2}$. Point $T$ is the mid-point of $P Q$. The ratio $Q U: U R=2: 1$, the ratio $R V: V S=3: 1$ and the ratio $S W: W P=4: 1$.
What is the area, in $\mathrm{cm}^{2}$, of quadrilateral $T U V W$ ?
A 66
B 67
C 68
D 69
E 70
SolUTION
B

We are told that $T$ is the mid-point of $P Q$. Hence $P T=\frac{1}{2} P Q$. Similarly, we are told that $S W: W P=4: 1$. Therefore $W P=\frac{1}{5} P S$. Hence $\frac{1}{2}(P T \times P W)=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{5}(P Q \times P S)$. It follows that the fraction of the area of square $P Q R S$ that lies in triangle $P T W$ is $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{5}$ or $\frac{1}{20}$. Similarly, the fraction of the square that lies in triangle $T Q U$ is $\frac{1}{2} \times \frac{1}{2} \times \frac{2}{3}$ or $\frac{1}{6}$, the fraction of the square that lies in triangle $U R V$ is $\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}$ or $\frac{1}{8}$ and the fraction of the square that lies in triangle $V S W$ is $\frac{1}{2} \times \frac{1}{4} \times \frac{4}{5}$ or $\frac{1}{10}$. Therefore the fraction of the square that lies outside quadrilateral $T U V W$ is $\frac{1}{20}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}=\frac{53}{120}$. Since we are told in the question that the area of square $P Q R S$ is $120 \mathrm{~cm}^{2}$, the area of quadrilateral $T U V W$, in $\mathrm{cm}^{2}$, is $\left(1-\frac{53}{120}\right) \times 120=67$.