One leg of a right triangle is 1 millimeter shorter than the longer leg and the hypotenuse is 1 millimeter longer than the longer leg.

Question

find the lengths of the sides of the triangle. 

Answer 1

Let l be the longer leg, then the shortest leg will be l-1

Use the Pythagorean Theorem: the hypotenuse of a right triangle equals the sum of the sides squared

 

l²  +  (l-1)²   =  (l +1)² 

 

l²   +  l² - 2l  +1  =  l²  + 2l  + 1

 

2l²  - 2l  +  1=  l²   +  2l   +  +1

 

l²  - 4l  =  0

 

l(l - 4)  = 0

 

l-4=0

 

l = 4, short leg =3, and hypotenuse = 5

 

The sides are 3, 4, and 5

Answer 2

x=the longer leg

x-1=the shorter leg

x+1=the hypotenuse

notice that x-1, x (the legs) and x+1 (the hypotenuse) are consecutive integers

a right triangle with consecutive integer measurements is a 3-4-5 right triangle

solving...

(x+1)²=x²+(x-1)²

x²+2x+1=x²+x²-2x+1

x²+x²-2x+1-x²-2x-1=0

combine similar terms

x²-4x=0

x(x-4)=0

x-4=0

x=4 mm

x+1=5 mm

x-1=3 mm

Answer 3

Since all sides of the triangle are being compared to the longer leg, we let the longer leg be a universal variable.

 

Let the longer leg = x

Let one leg = x - 1

Let hypotenuse = x + 1

 

Since we have a right triangle, use the Pythagorean theorem to find the dimensions of all three sides.

 

x² + (x - 1)² = (x + 1)²

 

Expand this out, we have

 

x² + (x² - 2x + 1) = x² + 2x + 1

x² + x² - 2x + 1 = x² + 2x + 1

2x² - 2x + 1 = x² + 2x + 1

 

Set the right side of the equation equal to zero by negating the signs of each term on the right side.

 

x² - 4x  = 0

x(x - 4) = 0

 

x = 0      and     x = 4

 

x cannot be zero because then you will have a dimension with no length.  

Accept x = 4.

 

 

Substitute x = 4 into the dimensions.

 

Longer leg = 4 mm

One leg = 3 mm

Hypotenuse = 5 mm

 

 

Notice that we have a 3, 4, 5 triangle.  This is known as the Pythagorean triples.  Also, the hypotenuse is the longest side of the triangle. Therefore, the dimensions are correct.

 

To check, we substitute the dimensions in the Pythagorean theorem.

 

4² + 3² = 5²

16 + 9 = 25

    25 = 25

 

This checks out.

Answer 4

Here's what we know:

a² + b² = c²

b = a + 1

c = b + 1

 

Now that we have some values, we can plug them into our formula:

a² + (a + 1)² = (b + 1)²

a² + (a + 1)² = ((a + 1) + 1)²

a² + (a + 1)² = (a + 2)²

a² + (a + 1)(a + 1) = (a + 2)(a + 2)

 

Now we FOIL everything out and then move everything to one side:

a² + a² + 2a + 1 = a² + 4a + 4

2a² + 2a + 1 = a² + 4a + 4

a² - 2a - 3 = 0

 

From that, we can easily factor:

(a - 3)(a + 1) = 0

a = either 3, or -1. But since we can't have a negative length, we can get rid of that and a = 3 millimeters.

This means b = 4 millimeters, and c = 5 millimeters.

 

Let's check our answer:

3² + 4² = 5²

9 + 16 = 25

25 = 25