Finding ratio of area of triangle and area of trapezium

Question

Can anyone help me with this? --> Image

I already answered a, b, c i) and c ii) but how am I supposed to do c iii)? I'm quite confused and I don't know how to solve this particular question.

Answer 1

Hint:

$\text{area}(NMCL)=\text{area}(ABC)-\text{area}(ANM)-\text{area}(NBL)$ and each of the l.h.s. area is a fraction of $\text{area}(ANM)$.

However, it's a parallelogram, not a trapezium.

Variant: $$\frac{\text{area}(ANM)}{\text{area}(NMCL)}=\frac{\text{area}(ANM)}{\text{area}(NBL)}\cdot\frac{\text{area}(NBL)}{\text{area}(NMCL)}$$ Observe that triangle $NBL$ and parallelogram $NMCL$ have the same height, hence the last ratio is $$\frac{\text{area}(NBL)}{\text{area}(NMCL)}=\frac{\frac12 BL}{NB}.$$

Answer 2

Working out part (c) in its entirety for better flow.

(i) $\frac{NM}{BC} = \frac 2{2+3} =\frac 25$

(ii) With similar figures, just square the side ratios to find the area ratios. So the ratio here is $(\frac 23)^2 = \frac 49$.

(iii) At least two ways to approach this.

The first (more complicated in my view) is to construct a relationship between the areas of $\triangle NBL$ and parallelogram $NMCL$ by observing that the base ratios $\frac {BL}{LC} = \frac 32$. The heights are identical. Since the formula for the area of a parallelogram is $base \times height$ and that of a triangle is $\frac 12 \times base \times height$, the ratio of the area of $\triangle NBL$ and parallelogram $NMCL$ is not $\frac 32$, but rather $\frac 34$ (to compensate for the extra factor of $\frac 12$). Finally, from part (ii)'s answer, the ratio of the areas of $\triangle ANM$ to $\triangle NBL$ is $\frac 49$, so the final answer is $\frac 49 \times \frac 34 = \frac 39 = \frac 13$.

The second (and much simpler in my view) approach is simply to observe that there are three similar triangles here, $\triangle ANM, \triangle NBL$ and $\triangle ABC$ with side ratios $2:3:5$. Therefore the area ratios are simply the squares of those, giving $4:9:25$. If we use arbitrary units for area, then $\triangle ANM$ will have area $4$ units, while the parallelogram $NMCL$ will have an area found by subtraction to be $25 - (4+9) = 12$ units. Hence the required ratio is $\frac 4{12} = \frac 13$, as before.