Question

In the figure, if $\frac{NT}{AB}=\frac{9}{5}$, and MB= 10 cm, then MN will be

• A. $18cm$
• B. $20cm$
• C. $30cm$
• D. $90cm$

Answer 1

The correct option is A $18cm$

$\frac{NT}{AB}=\frac{9}{5}$ and $MB=10cm$ [ Given ]

$\Rightarrow$ $\angle TBA+\angle MBA={180}^o$ [ Linear pair ]

$\Rightarrow$ ${90}^o+\angle MBA={180}^o$

$\Rightarrow$ $\angle MBA={180}^o-{95}^o$

$\therefore$ $\angle MBA={85}^o$

In $△MBA$ and $△MNT$

$\Rightarrow$ $\angle MBA=\angle MNT={85}^o$

$\Rightarrow$ $\angle AMB=\angle TMN$ [ Common angle ]

$\therefore$ $△MBA\sim \angle MNT$ [ By $AA$ similarity test ]

$\Rightarrow$ $\frac{MB}{MN}=\frac{AB}{NT}$ [ C.P.C.T ]

$\Rightarrow$ $\frac{10}{MN}=\frac{5}{9}$ [ $\because$ $\frac{NT}{AB}=\frac{9}{5}$ and $MB=10$ ]

$\Rightarrow$ $MN=\frac{90}{5}=18cm$