In parallelogram ABCD, E is a point in AB and DE meets diagonal AC at point F. If DF : FE = 5 : 3 and area of △ ADF is 60 cm²; find :

(i) area of △ ADE

(ii) if AE : EB = 4 : 5, find the area of △ ADB.

(iii) also, find area of parallelogram ABCD.

We know that,

Ratio of the area of triangles with same vertex and bases along the same line is equal to the ratio of their respective bases.

(i) △ ADF and △ AFE have same vertex A and their bases are on the same straight line DE.

$\therefore \dfrac{\text{Area of △ ADF}}{\text{Area of △ AFE}} = \dfrac{DF}{FE} \\[1em] \Rightarrow \dfrac{60}{\text{Area of △ AFE}} = \dfrac{5}{3} \\[1em] \Rightarrow \text{Area of △ AFE} = \dfrac{60 \times 3}{5} = \dfrac{180}{5} = 36 \text{ cm}^2.$

From figure,

Area of △ ADE = Area of △ ADF + Area of △ AFE = 60 + 36 = 96 cm².

Hence, area of △ ADE = 96 cm².

(ii) △ ADE and △ EDB have same vertex D and their bases are on the same straight line AB.

$\therefore \dfrac{\text{Area of △ ADE}}{\text{Area of △ EDB}} = \dfrac{AE}{EB} \\[1em] \Rightarrow \dfrac{96}{\text{Area of △ EDB}} = \dfrac{4}{5} \\[1em] \Rightarrow \text{Area of △ EDB} = \dfrac{96 \times 5}{4} = \dfrac{480}{4} = 120 \text{ cm}^2.$

From figure,

Area of △ ADB = Area of △ ADE + Area of △ EDB = 96 + 120 = 216 cm².

Hence, area of △ ADB = 216 cm².

(iii) We know that,

Diagonal of a parallelogram divides it into two triangles of equal area.

Area of parallelogram ABCD = 2 × Area of △ ADB = 2 × 216 = 432 cm².

Hence, area of || gm ABCD = 432 cm².