Respuesta :

AB is 2√137 in simplest form

or 23.4 rounded to the nearest tenth.

[tex] \huge \tt \color{pink}{A}\color{blue}{n}\color{red}{s}\color{green}{w}\color{grey}{e}\color{purple}{r }[/tex]

[tex] \large\underline{ \boxed{ \sf{✰\:Note }}}[/tex]

★ 1st let's know what is the given figure is and it's related concepts for solving !

➣ Given Triangle is a right angled triangle

➣ It is having 3sides let's know what are the name of these sides

➣ 1st AB is know as hypotenuse

➣ 2nd AC and is called Base of the triangle

➣ 3rd BC whích is know as perpendicular of the triangle

➣ Hypotenuse(H):-The side of a right triangle opposite the right angle.

➣ Perpendicular(P):- Exactly upright; extending in a straight line.

➣ Base(B):- it also known as the side opposite to hypotenuse

➣Perpendicular and base are know as the leg of right angled triangle

➣ We can easily find length of one missing side by using a theorem name as "Pythagorean theorem"

➣ Pythagorean theorem :- A mathematical theorem which states that the square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of those of the two other sides

★ Note :- The Pythagorean theorem only applies to right triangles.

[tex] \rule{70mm}{2.9pt}[/tex]

★ Writing this theorem mathematically ★

[tex] { \boxed{✫\underline{ \boxed{ \sf{Pythagorean \: theorem \: ⇒ {Hypotenuse }^2={ Base }^2+ {Height }^2}}}✫}}[/tex]

★ Here ★

➣ Base (AC)= 22cm

➣ Perpendicular (height) (BC)= 8cm

[tex] \rule{70mm}{2.9pt}[/tex]

✝ Assumption ✝

➣ let Hypotenuse ( AB ) = "x"

[tex] \boxed{ \rm{ \pink ➛AB^2= AC^2+BC^2}}[/tex]

[tex] \rule{70mm}{2.9pt}[/tex]

✝ let's substitute values ✝

[tex]\rm{ \pink ➛x^2=22 ^2+8^2} \\ \rm{ \pink ➛x^2 = 484+ {64} } \\ \rm{ \pink ➛ {x}^{2} = 548 } \\ \rm{ \pink ➛= {x} = \sqrt{548} } \\ \rm{ \pink ➛ x = 2\sqrt{137} \: or23.4} \\ [/tex]

[tex] \rule{70mm}{2.9pt}[/tex]

Hence Hypotenuse (AB) in the given triangle is of

[tex] { \boxed{✛\underline{ \boxed{ \sf{2\sqrt{137} \: or23.4\green✓}}}✛}}[/tex]

[tex] \rule{70mm}{2.9pt}[/tex]

Hope it helps !

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