Area ratios from vector ratios

Two facts about areas of triangles

Similar triangles: if corresponding lengths are in the ratio \(l_1:l_2\), the areas are in the ratio \(l_1^2:l_2^2\). Write the corresponding vertices in the same order before reading off the ratio.

Same height: triangles with a common vertex whose bases lie on one straight line have the same height, so their areas are in the ratio of their bases, even when the bases do not touch.

Left: \(JK\) is parallel to \(GH\) and \(FJ=\dfrac14FG\), so triangles \(FJK\) and \(FGH\) are similar and \(\dfrac{\text{area of triangle}\hspace{0.5em}FJK}{\text{area of triangle}\hspace{0.5em}FGH}=\left(\dfrac14\right)^2=\dfrac1{16}\). Right: triangles \(TUV\) and \(TWZ\) have the same height from \(T\) to the line \(UZ\), so their areas are in the ratio \(UV:WZ\).

Method with vectors

1. Get a ratio of lengths from a vector result. If \(\overrightarrow{KL}=\dfrac52\overrightarrow{MN}\), then \(KL:MN=5:2\); compare lengths and never divide one vector by another.
2. Choose the fact: parallel sides that make the angles match give similar triangles (square the ratio); a common vertex with bases on one line gives the same height (use the ratio as it is).
3. Chain ratios when one triangle links two others, and finish with the ratio or area that was asked for.

Example: \(C\) lies on \(AB\) with \(AC:CB=5:4\). Triangles \(OAC\) and \(OCB\) share the vertex \(O\) and their bases lie on \(AB\), so their areas are in the ratio \(5:4\). If the area of triangle \(OAB\) is \(27\ \text{cm}^2\), the area of triangle \(OAC\) is \(\dfrac59\times27=15\ \text{cm}^2\).

Chaining: if the areas of triangles \(T_1\) and \(T_2\) are in the ratio \(1:2\), and the areas of \(T_2\) and \(T_3\) are in the ratio \(4:9\), write both with the same number for \(T_2\): \(T_1:T_2=2:4\), so \(T_1:T_3=2:9\).

Check

Square the ratio only for similar triangles: \(FJ=\dfrac14FG\) gives an area ratio of \(\dfrac1{16}\), not \(\dfrac14\). For triangles with the same height, do not square.