SOLUTION: Hi, could someone please help with a word problem? The problem is: The hypotenuse of a right triangle is 5 m longer than one leg and 2 m longer than the other leg. Find the peri

Algebra ->  Pythagorean-theorem -> SOLUTION: Hi, could someone please help with a word problem? The problem is: The hypotenuse of a right triangle is 5 m longer than one leg and 2 m longer than the other leg. Find the peri      Log On


   

Question 1022855: Hi, could someone please help with a word problem? The problem is: The hypotenuse of a right triangle is 5 m longer than one leg and 2 m longer than the other leg. Find the perimeter of the right triangle. Thank you!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = the length in meters of
the shorter leg
Let +b+ = the length in meters of
the longer leg
Let +c+ = the length in meters of
the hypotenuse
------------------
(1) +c%5E2+=+a%5E2+%2B+b%5E2+
(2) +c+=+a+%2B+5+
(3) +c+=+b+%2B+2+
-----------------
(2) +a+=+c+-+5+
and
(3) +b+=+c+-+2+
----------------
Plug (2) and (3) into (1)
(1) +c%5E2+=+%28+c+-+5+%29%5E2+%2B+%28+c+-+2+%29%5E2+
(1) +c%5E2+=+c%5E2+-+10c+%2B+25+%2B+c%5E2+-+4c+%2B+4+
(1) +c%5E2+=+2c%5E2+-+14c+%2B+29+
(1) +c%5E2+-+14c+%2B+29+=+0+
Complete the square:
(1) +c%5E2+-+14c+%2B+%28+14%2F2+%29%5E2++=++-29+%2B+%2814%2F2+%29%5E2+
(1) +c%5E2+-+14c+%2B+49+=+-29+%2B+49+
(1) +c%5E2+-+14c+%2B+49+=+20+
(1) +%28+c+-+7+%29%5E2+=+20+
Take the square root of both sides
( note that I don't use the negative square root
since that gives me a negative result for +a+ )
(1) +c+-+7+=+sqrt%2820%29+
(1) +c+=+4.472%2B+7+
(1) +c+=+11.472+
and
(2) +a+=+c+-+5+
(2) +a+=+11.472+-+5+
(2) +a+=+6.472+
and
(3) +b+=+c+-+2+
(3) +b+=+11.472+-+2+
(3) +b+=+9.472+
--------------------
The perimeter is:
+11.472+%2B+6.472+%2B+9.472+=+27.416+
----------
check:
(1) +c%5E2+=+a%5E2+%2B+b%5E2+
(1 ) +11.472%5E2++=+6.472%5E2+%2B+9.472%5E2+
(1) +131.607+=+41.887+%2B+89.719+
(1) +131.607+=+131.606+
error due to rounding off
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