Question

The sides of a right angled triangle are xcm. x+1 cm and x+3 cm show that x square - 4x-8=0 and solve this equation to find the lengths of the sides of the triangle

Answer 1

${ \huge \mathfrak{ \red{ \underline{question}}}}$

The sides of a right angled triangle are x cm , ( x + 1 ) cm and ( x + 3 ) cm.

Show that

${ \bold{ {x}^{2} - 4x - 8 = 0}}$

and solve this equation to find the lengths of the sides of the triangle.

${ \huge{ \mathfrak{ \red{ \underline{solution}}}}}$

${ \bold{ \underline{ \orange{given}}}}$

sides of the right angled triangle are given....

let height , base and hypotenuse be x cm. , ( x + 1 ) cm and ( x + 3 ) cm respectively.....

${ \bold{ \underline{ \orange{concept \: used}}}}$

${ \blue{ \bold{pythagoras \: theorem}}}$

the Pythagorean theorem, also known as Pythagoras' theorem, is a fundamental relation in Euclidean geometry among the three sides of a right triangle. It states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides.

${ \boxed{ \red{ \bold{ {hypotenuse}^{2} = {height}^{2} + {base}^{2} }}}}$

${ \bold{ {(x + 3)}^{2} = {x}^{2} + {(x + 1)}^{2} }}$

• applying the algebraic identity to both the sides....

${ \bold{ \underline{ \pink{ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab}}}}$

${ \bold{ \implies{ {x}^{2} + {3}^{2} + 2(3)x = {x}^{2} + {x}^{2} + {1}^{2} + 2(1)x}}}$

${ \bold{ \implies{ {x}^{2} + 9 + 6x = 2 {x}^{2} + 2x + 1}}}$

${ \bold{ \implies{2 {x}^{2} - {x}^{2} + 2x - 6x + 1 - 9 = 0}}}$

${ \bold{ \implies{ \underline{ \red{ {x}^{2} - 4x - 8 = 0}}}}}$

now, solving this equation to find the lengths of the sides of the triangle.....

• using quadratic formula for this equation...

${ \boxed{ \bold{ \red{x = \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a} }}}}$

here,

• a = +1 , b = -4 and c = -8

• substituting the values in the formula....

${ \bold{x = \frac{ - ( - 4)± \sqrt{ {( - 4)}^{2} - 4(1)( - 8)} }{2 \times 1} }}$

${ \bold{ \implies{x = \frac{4± \sqrt{16 + 32} }{2} }}}$

${ \bold{ \implies{x = \frac{4± \sqrt{48} }{2} }}}$

${ \bold{ \implies{x = \frac{4±4 \sqrt{3} }{2} }}}$

${ \bold{ \implies{x = \frac{2(2 \: ± \: 2 \sqrt{3}) }{2} }}}$

${ \bold{ \implies{x = 2 ± \: 2 \sqrt{3} }}}$

• first root value of x....

${ \bold{x = 2 + 2 \sqrt{3} = 2 + 2(1.732)}}$

${ \bold{ \implies{x = 2 + 3.464 = 5.464}}}$

• second root value of x.....

${ \bold{x = 2 - 2 \sqrt{3} = 2 - 2(1.732)}}$

${ \bold{ \implies{x = 2 - 3.464 = - 1.464}}}$

• since,. the length of side can never be negative

therefore,

${ \boxed{ \red{ \bold{ \huge{x = 5.464 \: cm}}}}}$

• height = x cm = 5.464 cm

• base = ( x + 1 ) cm = ( 5.464 + 1 ) cm = 6.464 cm

• hypotenuse = ( x+3)cm = (5.464+3) cm = 8.464 cm