$$ABC$$ is a triangle. $$AB = 5 cm$$, $$AC = \surd41 cm$$ and $$BC = 8 cm.$$ $$AD$$ is perpendicular to $$BC$$. What is the area (in $$cm^2$$) of triangle $$ABD$$?

In the triangle ABC as AB=5 and AD is perpendicular to BD and so triangle ABD and triangle ADC are right angled triangles
AB=5 and so triangle ABD other sides may be 3 and 4 as BD=4 is not possible because if BD=4 then DC=4 the perpendicular is becoming also a bisector the the triangle ABC should be isosceles but it not the case so BD=3 and AD=4 and so DC=5
In triangle ADC also the condition satisfies and so area of the triangle ABD=(1/2)*3*4
=6 sq cm