The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point O. If BO : OD = 4 : 7, find :

(i) △AOD : △AOB

(ii) △AOB : △ACB

(iii) △DOC : △AOB

(iv) △ABD : △BOC

(i) Given,

⇒ BO : OD = 4 : 7

⇒ OD : OB = 7 : 4

Since,

△AOD and △AOB have common vertex at A and their bases OD and OB are along the same straight line.

$\therefore \dfrac{\text{Area of △AOD}}{\text{Area of △AOB}} = \dfrac{OD}{OB} = \dfrac{7}{4}$.

Hence, △AOD : △AOB = 7 : 4.

(ii) Since,

△AOB and △ACB have common vertex at A and their bases AO and AC are along the same straight line.

$\therefore \dfrac{\text{Area of △AOB}}{\text{Area of △ACB}} = \dfrac{AO}{AC} = \dfrac{4}{11}$

Hence, △AOB : △ACB = 4 : 11.

(iii) Given,

BO : OD = 4 : 7

In △AOB and △DOC

∠AOB = ∠COD [Vertically opposite angles are equal]

∠ABO = ∠ODC [Alternate angles are equal]

∴ △AOB ~ △DOC [By AA axiom]

By area theorem,

Ratio of area of two similar triangles is equal to the square of the ratio of the corresponding sides.

$\therefore \dfrac{\text{Area of △DOC}}{\text{Area of △AOB}} = \Big(\dfrac{OD}{OB}\Big)^2 \\[1em] = \Big(\dfrac{7}{4}\Big)^2 \\[1em] = \dfrac{49}{16} \\[1em] = 49 : 16.$

Hence, △DOC : △AOB = 49 : 16.

(iv) Given,

BO : OD = 4 : 7

$\Rightarrow \dfrac{BO}{OD} = \dfrac{4}{7} \\[1em] \Rightarrow \dfrac{OD}{BO} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{OD}{BO} + 1 = \dfrac{7}{4} + 1 \\[1em] \Rightarrow \dfrac{OD + BO}{BO} = \dfrac{7 + 4}{4} \\[1em] \Rightarrow \dfrac{BD}{BO} = \dfrac{11}{4}.$

Since,

△ABD and △BOC have common vertex at B and their bases BD and OB are along the same straight line.

$\therefore \dfrac{\text{Area of △ABD}}{\text{Area of △BOC}} = \dfrac{BD}{BO} = \dfrac{11}{4}.$

Hence, △ABD : △BOC = 11 : 4.