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Sagot :
Pythagoras Theorem:
In a right angle triangle, the sum of square of base and perpendicular is equal to the square of Hypotenuse .
Hypotenuse² = Perpendicular² + base²
In the given figure, we have:
Base = k
Hypotenuse = 10
Perpendicular = 6
Substitute the valus and solve for k,
[tex]\begin{gathered} \text{Hypotenuse}^2=Perpendicular^2+Base^2 \\ 10^2=6^2+k^2 \\ 100=36+k^2 \\ k^2=100-36 \\ k^2=64 \\ k=\sqrt[]{64} \\ k=8 \\ \text{Base, k = 8} \end{gathered}[/tex]The missing side is 8
Answer: 8
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