Answer:
Step-by-step explanation:
1 to find AB
in triangle ADB
here 4 and 2[tex]\sqrt{6}[/tex] are the legs of the right triangle so let them be a and b respectively.
AB is the hypotenuse (longest side ) and let it be x
using pythagoras theorem
a^2 + b^2 = c^2
4^2 + [tex](2\sqrt{6} )^2[/tex] = AB^2
16 + 4*6 = AB^2
16 + 24 = AB^2
40 = AB^2
[tex]\sqrt{40}[/tex] = AB
[tex]2\sqrt{10}[/tex] = AB
therefore AB =v[tex]2\sqrt{10}[/tex] .Hence proved .
In triangle ADC
here 6 and [tex]2\sqrt{6}[/tex] are the legs of the right angle so let them be a and b respectively. AC is the hypotenuse (longest side) so let it be c
usinfg pythagoras theorem
a^2 + b^2 = c^2
6^2 + [tex](2\sqrt{6} )^2[/tex] = AC^2
36 + 4*6 = AC^2
36 + 24 = AC^2
60 = AC^2
[tex]\sqrt{60}[/tex] = AC
[tex]2\sqrt{15}[/tex] = AC
To prove triangle ABC is a right angled triangle
AD and BD are the legs and AB is hypotenuse
according to pythagoras theorem to be the right angle triangle the sum of square of two smaller sides of a triangle must equal to the square of hypotenuse(longest side)
let two smaller sides be a and b respectively and c be hypotenuse
using pythagoras theorem
a^2 + b^2 = c^2
4^2 + [tex](2\sqrt{6} )^2[/tex] = [tex](2\sqrt{10} )^2[/tex]
16 +2^2*6 = 2^2*10
16 + 4*6 = 4*10
16 + 24 = 40
40 = 40
since both sides are equal it is proved that the triangle ABC is a right angled triangle.
Hope this helps u !!
