✰ Concept Used :-
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In this question, we can clearly observer that the diagram shows a right angled triangle. And, we have been provided with the value of base, and the value of hypotenuse, using the pythagoras theorem, now we can easily find out the value of the perpendicular i.e. the value of the side h. According to the pythagoras theorem, square of hypotenuse is equal to the sum of square of perpendicular and square of side respectively. Therefore, square of side is equal to the difference of square of hypotenuse and square of perpendicular.
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✰ Given Information :-
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- Hypotenuse = 22 ft.
- Base = 10 ft.
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✰ To Find :-
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- The value of side or the perpendicular
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✰ Formula Used :-
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[tex] \star \: \underline{ \boxed{ \purple { \sf {Side}^{2} = {Hypotenuse}^{2} - {Base}^{2} }}} \: \star [/tex]
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✰ Solution :-
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[tex]\sf \longrightarrow {Side}^{2} = {(22 \: ft)}^{2} - {(10 \: ft)}^{2} \: \: \: \\ \\ \\ \sf \longrightarrow {Side}^{2} = {484 \: ft}^{2} - {100 \: ft}^{2} \: \: \: \: \: \: \\ \\ \\ \sf \longrightarrow {Side}^{2} = {384 \: ft}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf \longrightarrow {Side}^{} = \sqrt{ {384 \: ft}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf \longrightarrow {Side}^{} = \underline{ \boxed{ \frak{ \green{19.60 \: ft}}}} \: \star \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ [/tex]
Thus, option B. 19.60 ft. is the correct option.
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[tex]\underline{\rule{230pt}{2pt}} \\ \\ [/tex]
