In the given figure, LM ∥ BC. If AB = 6 cm, AL = 2 cm and AC = 9 cm, calculate :

(i) the length of CM,

(ii) Find the value of $\dfrac{\text{ar(ΔALM)}}{\text{ar(trap. LBCM)}}$.

(i) Given,

AB = 6 cm

AL = 2 cm

LB = AB - AL = 6 - 2 = 4 cm

AC = AM + MC

AM = AC - MC

AM = 9 - MC

In ΔAML and ΔABC,

∠AML = ∠ACB [Corresponding angles are equal]

∠LAM = ∠BAC [Common angle]

ΔAML ∼ ΔABC [By AA similarity]

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AL}{LB} = \dfrac{AM}{MC} \\[1em] \Rightarrow \dfrac{2}{4} = \dfrac{9 - MC}{MC} \\[1em] \Rightarrow 2MC = 4(9 - MC) \\[1em] \Rightarrow 2MC = 36 - 4MC \\[1em] \Rightarrow 2MC + 4MC = 36 \\[1em] \Rightarrow 6MC = 36 \\[1em] \Rightarrow MC = \dfrac{36}{6} \\[1em] \Rightarrow MC = 6 \text{ cm.}$

Hence, CM = 6 cm.

(ii) Given

In ΔALM and ΔABC,

∠AML = ∠ACB [Corresponding angles are equal]

∠LAM = ∠BAC [Common angle]

ΔAML ∼ ΔABC [By AA similarity]

We know that,

The ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{ar(ΔALM)}}{\text{ar(ΔABC)}} = \Big(\dfrac{AL}{AB}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{ar(ΔALM)}}{\text{ar(ΔABC)}} = \Big(\dfrac{2}{6}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{ar(ΔALM)}}{\text{ar(ΔABC)}} = \dfrac{4}{36} \\[1em] \Rightarrow \dfrac{\text{ar(ΔALM)}}{\text{ar(ΔABC)}} = \dfrac{1}{9}.$

Let ar(ΔALM) = x, then ar(ΔABC) = 9x.

From figure,

ar(trap. LBCM) = ar(ΔABC) - ar(ΔALM)

= 9x - x

= 8x.

$\Rightarrow \dfrac{\text{ar(ΔALM)}}{\text{ar(trap. LBCM)}} = \dfrac{x}{8x} = \dfrac{1}{8}$.

Hence, $\dfrac{\text{ar(ΔALM)}}{\text{ar(trap. LBCM)}} = \dfrac{1}{8}$.