Question

1) in a right angled triangle one side is 2m shorter than the hypotenuse while the third side is one sixth of the sum of the other two.find the lengths of the sides and also the length of the perpendicular to the hypotenuse from the opposite vertex

Answer 1

hypo: x
one leg: x-2
other leg: (1/6)(x-2+x) = x/3 + 1/3

(x-2)^2 + (x/3 + 1/3)^2 = x^2
x^2-4x+4 + x^2/9 - 2x/9 + 1/9 = x^2
x^2/9 - 4x - 2x/9 + 37/9 = 0
multiply each term by 9
x^2 - 36x - 2x + 37 = 0
x^2 - 38x + 37 = 0

finish it up, it factors. Make sure you check to see if both answers are valid.

For the second part, you will have similar triangles, so use ratios

Answer 2

a = longer leg

b = shorter leg

c = hypotenuse

a = c - 2

b = ( a + c ) / 6

b = ( c - 2 + c ) / 6

b = ( 2 c - 2 ) / 6

b = 2 ( c - 1 ) / 2 ∙ 3

b = ( c - 1 ) / 3


a² + b² = c²

( c - 2 )² + [ ( c - 1 ) / 3 ]² = c²

c² - 2 ∙ c ∙ 2 + 2² + ( c - 1 )² / 3² = c²

c² - 4 c + 4 + ( c² - 2 ∙ c ∙ 1 + 1² ) / 9 = c²

c² - 4 c + 4 + ( c² - 2 c + 1 ) / 9 = c²

Multiply both sides by 9

9 c² - 36 c + 36 + c² - 2 c + 1 = 9 c²

9 c² + c² - 38 c + 37 = 9 c²

Subtract 9 c²

c² - 38 c + 37 = 0

The solutions are:

c = 1 , c = 37

The length cannot be c = 1 because then the length a = c - 2 = 1 - 1 = - 1 < 0

So:

c = 37 m

a = c - 2 = 37 - 2 = 35 m

b = ( a + c ) / 6 = ( 35 + 37 ) / 6 = 72 / 6 = 12 m

tan θ = b / a

tan θ = 12 / 35

sin θ = ± tan θ / √ ( 1 + tan θ² )

θ < 90° so sine and tangent are positive.

sin θ = tan θ / √ ( 1 + tan θ² )

sin θ = ( 12 / 35 ) / √ [ 1 + ( 12 / 35 )² ]

sin θ = ( 12 / 35 ) / √ [ 1 + ( 144 / 1225 ) ]

sin θ = ( 12 / 35 ) / √ ( 1225 / 1225 + 144 / 1225 )

sin θ = ( 12 / 35 ) / √ ( 1239 / 1225 )

sin θ = ( 12 / 35 ) / ( 37 / 35 )

sin θ = 12 / 37

sin θ = h / a

h = a ∙ sin θ

h = 35 ∙ 12 / 37 = 35 ∙ 12 / 37 = 420 / 37 = 11.35135135 m