Can anyone help me solve this problem?
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Given
[tex] \text{Base = 4} \\ \text{Perpendicular= 8}[/tex]
To Find
Hypotenuse of the triangle.
Assuming,
[tex] \text{Perpendicular = ‘a’}[/tex]
[tex] \text{Base = ‘b’}[/tex]
[tex] \text{Hypotenuse= ‘c’}[/tex]
[tex] \text{(c)² = (a)² + (b)²} \\ \\ \implies \text{ (c)² = (8)² + (4)²} \\ \\ \implies \text{(c)}^{2} = 64 + 16 \\ \\ \implies \text{(c)}^{2} = 80 \\ \\ \implies \text{(c)} = \sqrt{80} \\ \\ \implies\text{(c)} = 8.94[/tex]
[tex]\text{Rounding it to the nearest tenth = 8.9}[/tex]
[tex]\therefore \text{Hypotenuse of the triangle = 8.9}[/tex]
Option D= 8.9 units is the correct answer