In the figure (ii) given below, ABCD is a parallelogram. AM ⊥ DC and AN ⊥ CB. If AM = 6 cm, AN = 10 cm and the area of parallelogram ABCD is 45 cm², find

(i) AB

(ii) BC

(iii) area of △ADM : area of △ANB.

(i) Given, AM = 6 cm and AN = 10 cm and area of parallelogram ABCD is 45 cm².

Area of parallelogram = base x height = CD x AM = BC x AN.

$\therefore AM \times CD = 45 \\[1em] \Rightarrow 6 \times CD = 45 \\[1em] \Rightarrow CD = \dfrac{45}{6} \\[1em] \Rightarrow CD = \dfrac{15}{2} \\[1em] \Rightarrow CD = 7.5$

In parallelogram AB = CD = 7.5 cm.

Hence, the length of AB = 7.5 cm.

(ii) Given, AM = 6 cm and AN = 10 cm and area of parallelogram ABCD is 45 cm².

Area of parallelogram = base x height = CD x AM = BC x AN.

$\therefore AN \times BC = 45 \\[1em] \Rightarrow 10 \times BC = 45 \\[1em] \Rightarrow BC = \dfrac{45}{10} \\[1em] \Rightarrow BC = 4.5$

Hence, the length of BC = 4.5 cm.

(iii) Considering △ADM and △ABN,

∠ ADM = ∠ ABN (Opposite angles of a parallelogram are equal)
∠ AMD = ∠ ANB (Both angles are equal to 90°)

Hence, by AA axiom △ADM ~ △ANB.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ADM}}{\text{Area of △ANB}} = \dfrac{AM^2}{AN^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ADM}}{\text{Area of △ANB}} = \dfrac{6^2}{10^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ADM}}{\text{Area of △ANB}} = \dfrac{36}{100} \\[1em] \Rightarrow \dfrac{\text{Area of △ADM}}{\text{Area of △ANB}} = \dfrac{9}{25}$

Hence, the ratio of the area of △ADM : area of △ANB = 9 : 25.