Assertion (A): In △ABC, BD : DC = 1 : 2 and OA = OD

Area of △AOB : area of △ABC = 1 : 4

Reason (R):

Area of △AOB = $\dfrac{1}{2}\times \text{area of △ABD}$
=$\dfrac{1}{2}\times \dfrac{2}{3}\times \text{area of △ABC}$

=$\dfrac{1}{3}\times \text{area of △ABC}$

1. A is true, R is false.
2. A is false, R is true.
3. Both A and R are true.
4. Both A and R are false.

Both A and R are false.

Explanation

Δ ABD and Δ ABC have same height.

$\dfrac{Area(Δ ABD)}{Area(Δ ABC)} = \dfrac{\dfrac{1}{2} \times base$1 \times height}{\dfrac{1}{2} \times base2 \times height}\\[1em] = \dfrac{BD}{BC}\\[1em] = \dfrac{1}{1 + 2}\\[1em] = \dfrac{1}{3} ……………..(1)

△AOB and △ABD have the same height. Since OA = OD, the height of △AOB is half the height of △ABD.

$\dfrac{Area(Δ AOB)}{Area(Δ ABD)} = \dfrac{\dfrac{1}{2} \times base$1 \times height}{\dfrac{1}{2} \times base2 \times height}\\[1em] = \dfrac{OA}{AD}\\[1em] = \dfrac{1}{2} ……………….(2)

Combing the ratios,

$\dfrac{Area(Δ AOB)}{Area(Δ ABC)} = \dfrac{Area(Δ AOB)}{Area(Δ ABD)} \times \dfrac{Area(Δ ABD)}{Area(Δ ABC)}\\[1em]$

Substitute from equations (1) and (2):

$= \dfrac{1}{2} \times \dfrac{1}{3}\\[1em] = \dfrac{1}{6}$

∴ Assertion (A) is false.

In triangle AOB, OA = OD

Area of Δ AOB = $\dfrac{1}{2}$ x area of Δ ABD (From equation (2))

= $\dfrac{1}{2} \times \dfrac{1}{3}$ x area of Δ ABC (From equation (1))

= $\dfrac{1}{6}$ x area of Δ ABC

∴ Reason (R) is false.

Hence, both Assertion (A) and Reason (R) are false.