Answer : The value of m is, [tex]\sqrt{136}[/tex]
Step-by-step explanation :
Using Pythagoras theorem in ΔABC :
[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]
[tex](AC)^2=(AB)^2+(BC)^2[/tex]
[tex](9+8)^2=(AB)^2+(m)^2[/tex]
[tex](17)^2=(AB)^2+(m)^2[/tex] .............(1)
Using Pythagoras theorem in ΔBDC :
[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]
[tex](BC)^2=(BD)^2+(DC)^2[/tex]
[tex](m)^2=(BD)^2+(8)^2[/tex] ............(2)
Using Pythagoras theorem in ΔBDA :
[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]
[tex](AB)^2=(BD)^2+(DA)^2[/tex]
[tex](AB)^2=(BD)^2+(9)^2[/tex]
[tex](AB)^2-(9)^2=(BD)^2[/tex] ..........(3)
Substituting equation 3 and 2, we get:
[tex](m)^2=(BD)^2+(8)^2[/tex]
[tex](m)^2=(AB)^2-(9)^2+(8)^2[/tex]
[tex](m)^2=(AB)^2-81+64[/tex]
[tex](m)^2+17=(AB)^2[/tex] ...........(4)
Now put equation 4 in 1, we get:
[tex](17)^2=(AB)^2+(m)^2[/tex]
[tex](17)^2=(m)^2+17+(m)^2[/tex]
[tex](17)^2-17=2m^2[/tex]
[tex]289-17=2m^2[/tex]
[tex]272=2m^2[/tex]
[tex]m=\sqrt{\frac{272}{2}}[/tex]
[tex]m=\sqrt{136}[/tex]
Thus, the value of m is, [tex]\sqrt{136}[/tex]
