From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC.

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Solution

Given here is a right angled triangle. So, we can apply the Pythagoras theorem.
${AB}^{2} + {BC}^{2} = {AC}^{2} \Rightarrow 20^{2} + 21^{2} = {AC}^{2} \Rightarrow {AC}^{2} = 400 + 441 = 841 \Rightarrow AC = 29$
Thus, the length of hypotenuse is 29 units.
Perimeter of ∆ABC = AB + BC + CA = 20 + 21 + 29 = 70 units.

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From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC.

Given here is a right angled triangle. So, we can apply the Pythagoras theorem. AB2+BC2=AC2⇒202+212=AC2⇒AC2=400+441=841⇒AC=29 Thus, the length of hypotenuse is 29 units. Perimeter of ∆ABC = AB + BC + CA = 20 + 21 + 29 = 70 units.