SOLUTION: 14. The hypotenuse of a right triangle measures 29 cm. One leg is 1 cm shorter than the other. What are the lengths of the legs?

Question 697059: 14. The hypotenuse of a right triangle
measures 29 cm. One leg is 1 cm shorter
than the other. What are the lengths of
the legs?
Answer by RedemptiveMath(80) (Show Source): You can put this solution on YOUR website!
Since we are talking about right triangles, let us keep the Pythagorean theorem in mind. First, we are given that the hypotenuse of this right triangle is 29 cm. Then we are given uncertain values for the legs: The longer leg is 1 cm more in length than the shorter leg, or the shorter leg is 1 cm less in measure than the longer leg. We need to find the lengths of these legs, and we know we can find uncertain measures of legs by using the Pythagorean theorem:

a^2 + b^2 = c^2, where a and b are legs and c is the hypotenuse.

We have the formula fresh in our minds, but we need to plug values into this in order to make any progress. We know the hypotenuse is 29 cm, so we have one value:

a^2 + b^2 = 29^2.

Let us now examine the legs. I am going to explain the next steps by letting the shorter leg equal a variable only. We can do this by letting the longer leg equal the variable only, but let us do it this way.

We have two uncertain measures for the legs because we are not given specific values and we need to find what they equal. So, we know that they will both have variables in them. We need to find out how to write the measures of the legs in an algebraic way. Let us do the simpler variable step first by letting one of the legs equal the variable only. Let us let the shorter leg equal x. We can do this because we don't know anything about this leg in this case. Now we have to find the length of the longer leg. We know that the longer leg is 1 cm longer than the shorter, so we can make the measure of this leg equal x + 1. The only thing we know about this leg is that it is 1 cm greater than the shorter, and if the shorter leg equals x, then the longer must equal 1 more than that (x + 1). Now we can plug these values into the Pythagorean theorem and solve:

x^2 + (x+1)^2 = 29^2
x^2 + x^2 + 2x + 1 = 29^2 (FOIL on (x+1)^2)
2x^2 + 2x + 1 = 29^2 (combine like terms)
2x^2 + 2x + 1 = 841 (29^2)
2x^2 + 2x - 840 = 0 (subtract 841 from both sides in order to use the Zero-Product property)
2(x^2 + x - 420) = 0 (we can factor 2 out from the left side in order to help us factor more easily)
2(x + 21)(x-20) = 0 (factor using Zero-Product)
x = 20 or x = -21.

We didn't have to factor the 2 out from (2x^2 + 2x - 840), but it makes things easier if we can factor out a whole number from the polynomial before we begin using Zero-Product. We are given the normal two-answer solution, but we need to know which x to use. Since we are talking about measurement, it is in order to use the positive answer for x. So, we find x to be 20. Now we just plug this in for what we said the legs equaled:

Shorter leg = x = 20 cm.
Longer leg = x + 1 = 21 cm.
Hypotenuse = 29 cm.

There are your three side measurements.

It is possible to do it how the problem gives the information. That is, it would give equivalent answers if we let the longer leg equal x and the shorter leg equal x - 1 (1 cm shorter than the longer leg). Just to check, I'll do it this way below:

x^2 + (x-1)^2 = 29^2
x^2 + x^2 - 2x + 1 = 29^2
x^2 + x^2 - 2x + 1 = 841
2x^2 - 2x + 1 = 841
2x^2 - 2x - 840 = 0
2(x^2 - x - 420) = 0
2(x - 21)(x + 20) = 0
x = 21 or x = -20.

Longer leg = x = 21 cm
Shorter leg = x - 1 = 20 cm
Hypotenuse = 29 cm.

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