Respuesta :

Answer:

assume [tex]13 = z \\ 8 =x \\ bc = y[/tex]

by phytagorean theorem

[tex]{y}^{2} + {x}^{2} = z^{2} [/tex]

so

[tex]bc = y = \sqrt{ {z}^{2} - {x}^{2} } = \sqrt{169 - 64} = \sqrt{105} [/tex]

[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)= 8cm
  • ➣ Hypotenuse (BA)= 13cm

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

✝ Assumption ✝

  • ➣ let perpendicular/length ( BC ) = "x"

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

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

✝ let's substitute values ✝

[tex]\rm{ \pink ➛13^2= 8^2+x^2} \\ \rm{ \pink ➛169 = 64 + {x}^{2} } \\ \rm{ \pink ➛169 - 64 = {x}^{2} } \\ \rm{ \pink ➛105 = {x}^{2}} \\ \rm{ \pink ➛ \sqrt{105} \: or \: 10.2469 = x} \\ [/tex]

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

Hence length (BC) in the given triangle is of

[tex] { \boxed{✛\underline{ \boxed{ \sf{\sqrt{105} cm\: or \: 10.2469cm\green✓}}}✛}}[/tex]

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

Hope it helps !

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