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In the given figure, $\text{AB \| DC}$. If $\text{OB = 3OD}$ and $\text{CD = 1.8 cm}$, then find the length $\text{AB}$.
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Solution:
$\text{DC \| AB}$
$\triangle \text{OCD} \sim \triangle \text{OAB}$ (by AA similarity) (
frac{1}{2} Mark)
$\Rightarrow \frac{\text{OD}}{\text{OB}} = \frac{\text{CD}}{\text{AB}}$ (
frac{1}{2} Mark)
$\Rightarrow \frac{1}{3} = \frac{1.8}{\text{AB}}$ (
frac{1}{2} Mark)
$\Rightarrow \text{AB = 5.4 cm}$ (
frac{1}{2} Mark)
$\text{DC \| AB}$
$\triangle \text{OCD} \sim \triangle \text{OAB}$ (by AA similarity) (
frac{1}{2} Mark)
$\Rightarrow \frac{\text{OD}}{\text{OB}} = \frac{\text{CD}}{\text{AB}}$ (
frac{1}{2} Mark)
$\Rightarrow \frac{1}{3} = \frac{1.8}{\text{AB}}$ (
frac{1}{2} Mark)
$\Rightarrow \text{AB = 5.4 cm}$ (
frac{1}{2} Mark)