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
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
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
Let's denote the lengths of the sides of the right-angled triangle as follows:
Hypotenuse: x
One side: x - 2
Third side: (x + x - 2) / 6 = (2x - 2) / 6 = (x - 1) / 3
According to the Pythagorean theorem, in a right-angled triangle, the sum of the squares of the two shorter sides is equal to the square of the hypotenuse.
Using this information, we can set up the following equation:
(x - 2)^2 + [(x - 1) / 3]^2 = x^2
Expanding and simplifying the equation:
(x^2 - 4x + 4) + (1/9)(x^2 - 2x + 1) = x^2
9(x^2 - 4x + 4) + (x^2 - 2x + 1) = 9x^2
9x^2 - 36x + 36 + x^2 - 2x + 1 = 9x^2
Combining like terms:
10x^2 - 38x + 37 = 9x^2
x^2 - 38x + 37 = 0
This quadratic equation can be factored:
(x - 1)(x - 37) = 0
From here, we have two possible solutions:
x - 1 = 0 --> x = 1
x - 37 = 0 --> x = 37
Since the problem states that the hypotenuse is longer than one of the other sides, we can eliminate x = 1 as a valid solution.
Therefore, x = 37.
Substituting this value back into the original equations, we can find the lengths of the sides:
One side: x - 2 = 37 - 2 = 35
Third side: (x - 1) / 3 = (37 - 1) / 3 = 36 / 3 = 12
Finally, to find the length of the perpendicular from the opposite vertex to the hypotenuse, we can apply the formula:
Perpendicular = (One side * Third side) / Hypotenuse
Perpendicular = (35 * 12) / 37 = 420 / 37 ≈ 11.35 meters
Therefore, the lengths of the sides of the right-angled triangle are 35, 12, and the length of the perpendicular to the hypotenuse from the opposite vertex is approximately 11.35 meters.
To solve this problem, we'll need to use the Pythagorean theorem and apply some algebraic reasoning.
Let's assume the hypotenuse of the right-angled triangle is x meters.
According to the problem, one side is 2 meters shorter than the hypotenuse. Therefore, the length of this side would be (x - 2) meters.
The third side is described as one-sixth of the sum of the other two sides. So, the length of this side would be ((x - 2) + x) / 6 meters.
Now, let's apply the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.
Therefore, we have the equation:
(x - 2)^2 + [((x - 2) + x)/6]^2 = x^2
Expanding and simplifying this equation:
(x^2 - 4x + 4) + [(2x - 2)/6]^2 = x^2
Simplifying further, we have:
x^2 - 4x + 4 + (4x^2 - 8x + 4)/36 = x^2
Multiplying through by 36 to get rid of the fractions:
36x^2 - 144x + 144 + 4x^2 - 8x + 4 = 36x^2
Combining like terms and simplifying:
40x^2 - 152x + 148 = 36x^2
4x^2 - 152x + 148 = 0
Now, we can solve this quadratic equation for x. We can either factor it or use the quadratic formula.
Using the quadratic formula:
x = [-(-152) ± sqrt((-152)^2 - 4(4)(148))] / (2 * 4)
x = [152 ± sqrt(23104 - 2368)] / 8
x = [152 ± sqrt(20736)] / 8
x = [152 ± 144] / 8
Now, we have two possible solutions:
1) When x = (152 + 144) / 8 = 296 / 8 = 37 meters
2) When x = (152 - 144) / 8 = 8 / 8 = 1 meter
Since the hypotenuse of a right-angled triangle cannot be shorter than either of the other two sides, we discard the second solution (x = 1 meter).
Therefore, the lengths of the sides of the right-angled triangle are:
Hypotenuse = 37 meters
One side = 35 meters (37 - 2)
Third side = 6.33 meters [((37 - 2) + 37) / 6]
To find the length of the perpendicular to the hypotenuse from the opposite vertex, we can use similar triangles. The perpendicular will be a height of the triangle, bisecting the hypotenuse.
By considering the smaller right-angled triangle that is similar to the original triangle, formed by the perpendicular and two smaller sides, we can use the Pythagorean theorem again.
Let the length of the perpendicular be h.
Using the smaller right-angled triangle, we have:
h^2 + (35/2)^2 = (37/2)^2
h^2 + 1225/4 = 1369/4
h^2 = 144/4
h^2 = 36
h = sqrt(36)
h = 6 meters
Hence, the length of the perpendicular to the hypotenuse from the opposite vertex is 6 meters.