ABCD and BCFE are parallelograms. If area of triangle EBC = 480 cm², AB = 30 cm and BC = 40 cm; Calculate :

(i) area of parallelogram ABCD;

(ii) area of the parallelogram BCFE;

(iii) length of altitude from A on CD;

(iv) area of triangle ECF.

(i) We know that,

The area of a triangle is half that of a parallelogram on the same base and between the same parallels.

From figure,

△ EBC and || gm ABCD lie on same base BC and between same parallel lines AD and BC.

∴ Area of △ EBC = $\dfrac{1}{2} \times$ Area of || gm ABCD

⇒ 480 = $\dfrac{1}{2} \times$ Area of || gm ABCD

⇒ Area of || gm ABCD = 2 × 480 = 960 cm².

Hence, area of || gm ABCD = 960 cm².

(ii) We know that,

Parallelograms on equal bases and between the same parallels are equal in area.

From figure,

Parallelogram ABCD and BCFE lie on same base BC and between same parallel lines AF and BC.

∴ Area of || gm BCFE = Area of || ABCD = 960 cm².

Hence, area of || gm BCFE = 960 cm².

(iii) We know that,

The area of a triangle is half that of a parallelogram on the same base and between the same parallels.

From figure,

△ ACD and || gm ABCD lie on same base AD and between same parallel lines AD and BC.

∴ Area of △ ACD = $\dfrac{1}{2} \times$ Area of || gm ABCD

⇒ Area of △ ACD = $\dfrac{1}{2} \times 960$ = 480 cm².

Since, opposite sides of parallelogram are equal.

∴ CD = AB = 30 cm.

By formula,

⇒ Area of triangle = $\dfrac{1}{2}$ × base × height

⇒ Area of triangle ACD = $\dfrac{1}{2} \times CD \times AP$

⇒ 480 = $\dfrac{1}{2} \times 30 \times AP$

⇒ AP = $\dfrac{480 \times 2}{30}$ = 32 cm.

Hence, length of altitude from A on CD = 32 cm.

(iv) We know that,

The area of a triangle is half that of a parallelogram on the same base and between the same parallels.

From figure,

△ EFC and || gm BCFE lie on same base EF and between same parallel lines EF and BC.

∴ Area of △ EFC = $\dfrac{1}{2} \times$ Area of || gm BCFE

⇒ Area of △ EFC = $\dfrac{1}{2} \times 960$ = 480 cm².

Hence, area of triangle ECF = 480 cm².