A right triangle has a hypotenuse that is 3 feet longer than one leg. The other leg is 4 feet.

Question

Please explain how you solved it.

Answer 1

Legs:

 

a = 4 ft

b = ?

 

Hypotenuse:

 

c = 3+b

 

c² = a²+b² 

 

(3+b)² = 4²+b²

 

9+6b+b² = 16+b²

 

6b = 16-9 = 7

 

b = 7/6 ft = 1.167 ft

 

c = 1.167+3 ft = 4.167 ft

Answer 2

Assuming the length of the hypotenuse (in feet) is the quantity to be solved for, you can use the Pythagorean theorem and the knowledge that the hypotenuse is three feet longer than one leg.  Then:

 

h² = (leg 1)²+(leg 2)²

 

Since the hypotenuse is 3 feet longer than one leg -- let's say, leg 1 -- then leg 1 = h-3 feet.  And leg 2, then, is 4 feet.  So the Pythagorean expression above becomes:

 

h² = (h-3)²+4² = (h-3)(h-3) + 16 = h²-6h+9 + 16

 

Thus:

 

h² = h² - 6h +25

 

After some algebra, this becomes:

 

6h = 25

 

which you can solve for h, the length of hypotenuse in feet.

 

Hope this helps!  Just let me know if you have any questions about this or related problems.