Question

If, in a $\Delta ABC,b=12$ units, $c=5$ units and $\Delta =30sq.units$ then the distance between vertex $A$ and incentre of the triangle is equal to

• A. 2 units
• B. $2\sqrt{2}$ units
• C. $\sqrt{2}$ units
• D. $3\sqrt{2}$ units

Answer 1

The correct option is B $2\sqrt{2}$ units

$△=\frac{1}{2}bc\sin A$

Therefore

$30=\frac{12\times 5}{2}\sin A$

Therefore

$\sin A=1$

Hence

$A=\frac{\pi }{2}$

Hence it is a right angles triangle

Therefore

$a^2=b^2+c^2$

$=144+25$

$=169$

Hence

$a=13$

Now

$AI=rcosec\frac{A}{2}$

$=\frac{△}{s}.cosec\frac{A}{2}$

$=\frac{30}{\frac{12+13+5}{2}}.cosec{45}^0$

$=2\sqrt{2}$