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One side of a right triangle is 2 cm shorter than the hypotenuse and 7 cm longer than the third side. find the lengths of the sides of the triangle.

Respuesta :

The right angles triangle has three sides including the hypotenuse which is opposite to the right angle.
If the hypotenuse - H
Then one side is - H - 2
Third side is 7 cm shorter then H-2
Therefore it’s H -2-7 = H -9
According to Pythagoras theorem
Square of hypotenuse = sum of the squares of the other 2 sides
H² = (H-9)² + (H-2)²
H² = H² -18H + 81 + H² -4H + 4
0 =H² -22H +85
This is a quadratic equation
0 = H² - 17H -5H +85
0 = H (H - 17) -5(H-17)
(H-17) (H-5) =0
H -17=0 or H-5=0
H could be 17 cm or 5 cm
If H =5 cm then
Side 1 -H-2 = 3 cm
And side 2 - H-9 = -4 cm
Since lengths cannot be negative values
H isn’t 5 cm
Therefore H =17 cm
1 side is H-2 = 15 cm
2nd side H -9 = 8 cm
Three sides are 17 cm, 15 cm and 8 cm

The sides will be 17, 15, and 8.

Triangle

It is a polygon of three sides and the sum of the internal angle of a triangle is 180 degrees.

Given

One side of a right triangle is 2 cm shorter than the hypotenuse

And 7 cm longer than the third side.

To find

The lengths of the sides of the triangle.

How do find the lengths of the sides of the triangle?

Let the hypotenuse is H

Then the One side of a right triangle is 2 cm shorter than the hypotenuse which means H - 2

And another side is 7 cm longer than the third side which means H -2 -7 that becomes H - 9

Then according to Pythagoras theorem

[tex]\rm H^{2} = P^{2} +B^{2}[/tex]

Then

[tex]\begin{aligned} \rm H^{2} &= (H-2)^{2} +(H-9)^{2}\\\rm H^{2} &= H^{2} -4H + 4+ H^{2} -18H +81\\ H^{2} -22H + 85 &= 0 \\(H-17)(H-5) &=0\\\end{aligned}[/tex]

Hence the value of H is 17 and 5.

5 is neglected because H - 9 is negative.

So the sides will be 17, 15, and 8.

More about the triangle link is given below.

https://brainly.com/question/25813512

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