Question

in the diagram, pqrs is a trapezium in which ps is parallel to qr. the diagonals pr and qs meet at t so that qt = 2 cm and ts = 1 cm. given that the area of triangle trs is 6 cm², find the area of (i) triangle qpt. 2 (ii) triangle qps. 2

Explanation:

Step1: Prove similarity of triangles

Since $PS\parallel QR$, $\triangle QPT\sim\triangle RST$ by the AA (angle - angle) similarity criterion (alternate - interior angles are equal). The ratio of the corresponding side lengths is $\frac{QT}{TS}=\frac{2}{1}$.

Step2: Use the ratio of areas of similar triangles

For two similar triangles with a side - length ratio of $k$, the ratio of their areas is $k^{2}$. Here, $k = 2$, so $\frac{Area(\triangle QPT)}{Area(\triangle RTS)}=k^{2}=4$. Given $Area(\triangle RTS)=6\ cm^{2}$, then $Area(\triangle QPT)=4\times6 = 24\ cm^{2}$.

Step3: Find the area of $\triangle QPS$

We know that $\triangle QPT$ and $\triangle PTS$ share the same height (the perpendicular distance between $PS$ and the line containing $Q$). The ratio of their bases is $\frac{QT}{TS}=\frac{2}{1}$, so $\frac{Area(\triangle QPT)}{Area(\triangle PTS)}=\frac{QT}{TS}=2$. Since $Area(\triangle QPT) = 24\ cm^{2}$, then $Area(\triangle PTS)=\frac{24}{2}=12\ cm^{2}$. Then $Area(\triangle QPS)=Area(\triangle QPT)+Area(\triangle PTS)=24 + 12=36\ cm^{2}$.

Answer:

(i) $24\ cm^{2}$
(ii) $36\ cm^{2}$