• Question 1 :-
We see that in a right angled triangle two sides are 8cm and 14cm . We can find the third side using Pythagoras Theorem .
[tex]\tt \to Hypotenuse^2 = Perpendicular^2+Base^2\\\\\tt \to h^2 = (8)^2+(14) \\\\\to \tt h^2 = 64 + 196 \\\\\tt \to h^2 = 260 \\\\\tt \to h=\sqrt{260} \\\\\tt \to \boxed{\orange{\tt h = 16.12 }}[/tex]
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• Question 2 :-
[tex]\tt \to Hypotenuse^2 = Perpendicular^2+Base^2\\\\\tt \to h^2 = (9)^2+(7)^2 \\\\\to\tt h^2 = 81 + 49 \\\\\tt \to h^2 = 130 \\\\\tt \to h=\sqrt{130} \\\\\tt \to \boxed{\orange{\tt h = 11.40}}[/tex]
