*ABCD* is a rectangle with *O* as any point in its interior. If ar (Δ*AOD*) = 3 cm^{2} and ar (Δ*ABOC*) = 6 cm^{2}, then area of rectangle *ABCD* is

#### Options

9 cm

^{2}12 cm

^{2}15 cm

^{2}18 cm

^{2}

#### Solution

**Given:** A rectangle ABCD , O is a point in the interior of the rectangles such that

(1) ar (ΔAOB) = 3 cm^{2}

(2) ar (ΔBOC) = 6 cm^{2}

**To find:** ar (rect.ABCD)

**Construction:** Draw a line LM passing through O and parallel to AD and BC.

**Calculation:** We know that **,” If a triangle and a parallelogram are on the same base and between the same parallels the area of the triangle is equal to half the area of the parallelogram”**

Here we can see that ΔAOD and rectangle AMLD are on the same base AD and between the same parallels AD and LM.

Hence ,

ar (Δ AOD) = `1/2` (rect . ALMD)

ar (rect ALMD )= 2 ar (ΔAOD)

ar (rect . ALMD) = 2(3)

ar (rect . ALMD ) = 6 cm^{2 }.....................(1)

Similarly, we can see that ΔBOC and rectangle BCLM are on the same base BC and between the same parallels BC and LM

Hence,

ar(ΔBOC ) = `1/2` ar (rect . BCLM)

ar (rect BCLM) = 2ar (ΔBOC)

ar (rect .aBCLM) = 12 cm^{2}

ar (rect . bclm) = 12 cm^{2} .................(2)

We known that

ar (rect . ABCD) = ar (rect . ALMD) + ar (rect . BCLM)

ar (rect . ABCD) = 6 +12

ar (rect . ABCD) = 18 cm^{2}