The base and altitude of a right angled triangle are 12 cm and 5 cm respectively. The perpendicular distance of its hypotenuse from the opposite vertex is:
Solution
In $\Delta$ABC,
AC $=\sqrt{12^2 + 5^2} = \sqrt{144 + 25}$
$=\sqrt{169} = 13$
Let AO $=x$. Then
OC $=13-x$
OB$^2 = 5^2-x^2= 25-x^2$ ... (i)
OB$^2 =12^2 - (13-x)^2$
$=144 - 169 - x^2 + 26x$ ...(ii)
From (i) and (ii)
$25-x^2 = -25 - x^2 + 26x$
$\Rightarrow 26x = 50$
$\Rightarrow x = \dfrac{50}{26} = \dfrac{25}{13}$
$\therefore$ OB$^2 = 25 - x^2$
$= 25-\left(\dfrac{25}{13}\right)^2$
OB$^2 = 25\left(1-\dfrac{25}{169}\right) = 25\times \dfrac{144}{169}$
OB $= \sqrt{\frac{25\times 144}{169}} = \dfrac{5\times 12}{13} = \dfrac{60}{13}$
$=4\dfrac{8}{13}$ cm