Answer:
[tex] \boxed{ \bold{24}}[/tex]
Step-by-step explanation:
[tex] \mathsf{given}[/tex]
[tex] \mathsf{hypotenuse(h) = 25}[/tex]
[tex] \sf{perpendicular (p) = 7}[/tex]
[tex] \sf{base(b) = }[/tex]?
Now, Using Pythagoras theorem
[tex] \sf{{h}^{2} = {p}^{2} + {b}^{2} }[/tex]
plug the values
⇒[tex] \sf{ {25}^{2} = {7}^{2} + {b}^{2} }[/tex]
Evaluate the power
⇒[tex] \sf{625 = 49 + {b}^{2} }[/tex]
Swap the sides of the equation
⇒[tex] \sf{49 + {b}^{2} = 625}[/tex]
Move constant to right hand side and change it's sign
⇒[tex] \sf{ {b}^{2} = 625 - 49}[/tex]
Calculate the difference
⇒[tex] \sf{ {b}^{2} = 576}[/tex]
Squaring on both sides
⇒[tex] \sf{b = 24}[/tex]
Hope I helped!
Best regards!
