In a right triangle, the hypotenuse is 8 dm longer than one leg and 4 dm longer than the other leg. Find the lengths of the three sides.

Question

 the hypotenuse is 8 dm longer than one leg and 4 dm longer than the other leg. Find the lengths of the three sides.

Answer 1

Both legs are being compared to the hypotenuse.

 

Let the hypotenuse = x

Let one leg = x - 8

Let the other leg = x - 4

 

Let me explain the logic behind these variables.  If the hypotenuse is 8 dm longer than one leg, then one leg is 8 dm shorter than the hypotenuse.  The same applies to the other leg.

 

 

Using these variables, we can set an equation.  Since we have a right triangle, we will apply the Pythagorean theorem.

 

(x - 8)² + (x - 4)² = x²

 

 

Solve for x from this equation.

 

Expanding, we get

 

(x - 8)(x - 8) + (x - 4)(x - 4) = x²

 

(x² - 16x + 64) + (x² - 8x + 16) = x²

 

2x - 24x + 80 = x²

 

 

Subtract x² on both sides of the equation.

 

x² - 24x + 80 = 0

 

 

We now have a quadratic equation.

 

 

We can factor using FOIL.

 

(x - 20)(x - 4) = 0

 

 

Set the factors equal to zero and solve for x from each part.

 

x - 20 = 0                  and                 x - 4 = 0

 

x = 20                       and                 x = 4

 

 

Since plugging in x=4   makes one of our legs a zero length, we accept   x=20.  Therefore,

 

Hypotenuse = 20 dm

One leg = 12 dm

Other leg = 16 dm

Answer 2

Let the length of the hypotenuse be h, and the lengths of the legs be a and b. 

Let's run the problem through the English-to-math converter:

 

"The hypotenuse"  "is" "8dm longer than one leg"

             h              =       a + 8

Or, a = h - 8.

 

Also, "The hypotenuse" "is" "4dm longer than the other leg"

                       h           =       b + 4

Or, b = h-4.

 

From the Pythagorean Theorem,

a² + b² = h².  By substitution,

(h-8)² + (h-4)² = h²

 

Use a little bit of the old (a-b)² = a² - 2ab + b², gather like terms on one side of the = and a big fat 0 on the other (standard procedure for solving quadratics), and you're on your way to finding a value for h.  From there, you can use a = h - 8 and b = h - 4 to find the lengths of the legs.  And don't forget to include the units (dm) in your answer.

Answer 3

Here is how to do many problems of this type.

 

1.  Draw the right triangle. 

2.  Label its side lengths, using a variable (x) for the basic unknown and using other expressions related to it. In this case, 'one leg' is x, the hypotenuse is x+8 and the remaining side (leg) is x+4, since the hypotenuse is 4 dm longer than the other leg.

3. Hypotenuse squared = leg squared + other leg squared; substitute the appropriate expressions into this equation.

4. Use Algebra to simplify.  Remember, for example, that (x+8)² = x² +16x + 64 (squaring a binomial ALWAYS gives a trinomial, with middle term 2 times the product of the two parts of the binomial).

5. You will get a quadratic equation; put all terms on one side, obeying proper handling of terms & their signs, and solve either by factoring (if lucky) or by using the Quadratic Formula, which is a very important thing to know.