Question

Work out length of BC

Answer 1

assume
`13 = z \\ 8 =x \\ bc = y`

by phytagorean theorem

`{y}^(2) + {x}^(2) = z^(2)`

so

`bc = y = \sqrt{ {z}^(2) - {x}^(2) } = √(169 - 64) = √(105)`

Answer 2

`\huge \tt \color{pink}{A}\color{blue}{n}\color{red}{s}\color{green}{w}\color{grey}{e}\color{purple}{r }`

`\large\underline{ \boxed{ \sf{✰\:Note }}}`

★ 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.

`\rule{70mm}{2.9pt}`

★ Writing this theorem mathematically ★

`{ \boxed{✫\underline{ \boxed{ \sf{Pythagorean \: theorem \: ⇒ {Hypotenuse }^2={ Base }^2+ {Height }^2}}}✫}}`

★ Here ★

• ➣ Base (AC)= 8cm
• ➣ Hypotenuse (BA)= 13cm

`\rule{70mm}{2.9pt}`

✝ Assumption ✝

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

`\boxed{ \rm{ \pink ➛BA^2= AC^2+BC^2}}`

`\rule{70mm}{2.9pt}`

✝ let's substitute values ✝

`\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 ➛ √(105) \: or \: 10.2469 = x} \\`

`\rule{70mm}{2.9pt}`

Hence length (BC) in the given triangle is of

`{ \boxed{✛\underline{ \boxed{ \sf{√(105) cm\: or \: 10.2469cm\green✓}}}✛}}`

`\rule{70mm}{2.9pt}`

Hope it helps !