Please help with the following equation: The hypotenuse of a right triangle is 1cm more than twice the shorter leg, and the longer leg is 9cm less than three times the shorter leg. Find the lengths of the three sides of the triangle. Thanks, Paul
Side lengths of a triangle
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>The hypotenuse of a right triangle is 1cm more than twice the shorter leg, and the longer leg is 9cm less than three times the shorter leg.
Name things. Let the short leg be called x. Now, how would write a mathematical expression that says "the hypotenuse is 1 more than twice the short leg (x)"? Now write the expression for the long leg that is "9 less than three times x."
Your next step is to use the Pythagorean Theorem to build your equation. Solve that for x. You should get two possibilities for x, namely 5/3 and 8. Then you can find the values of the other sides. Check to make sure they are all realistic values.
Name things. Let the short leg be called x. Now, how would write a mathematical expression that says "the hypotenuse is 1 more than twice the short leg (x)"? Now write the expression for the long leg that is "9 less than three times x."
Your next step is to use the Pythagorean Theorem to build your equation. Solve that for x. You should get two possibilities for x, namely 5/3 and 8. Then you can find the values of the other sides. Check to make sure they are all realistic values.
pka
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Do you know the Pythagorean theorem for right triangles?
\(\displaystyle h^2 = s^2 + l^2\)
If s is the length of the shortest side, then the hypotenuse is 1+2s and the longer leg is 3s-9.
So by the Pythagorean theorem \(\displaystyle (1+2s)^2 = s^2 + (3s-9)^2\)
You should find two solutions for s.
\(\displaystyle h^2 = s^2 + l^2\)
If s is the length of the shortest side, then the hypotenuse is 1+2s and the longer leg is 3s-9.
So by the Pythagorean theorem \(\displaystyle (1+2s)^2 = s^2 + (3s-9)^2\)
You should find two solutions for s.
Loren said:
The answers were exactly what i got as well. Here are my steps -
Let x be the short leg:
Short leg = x, Longer leg = 3x -9, Hypotenuse = 2x + 1
1) (3x - 9)^2 + x^2 = (2x + 1)^2
2) 9x^2 - 54x + 81 + x^2 = 4x^2 + 4x + 1
3) 10x^2 - 54x + 81 = 4x^2 + 4x + 1
4) 6x^2 + 80 = 58x
5) 6x^2 - 58x + 80 =0
6) Factorise the "2" = 2 (3x^2 - 29x + 40) =0
7) I used the cross method to factorise the equation so it becomes: 2 ( 3x - 5)(x - 8) = 0
8) Therefore x = 5/3 or x = 8.
I hope this helped you understand clearer.