**Right Triangle: Finding the Dimensions Knowing Only Area and Perimeter: finding a general formula**

Problem: Given the
area of a right triangle (R) and perimeter (P), find the dimensions of the
right triangle (a, b, and the hypotenuse c).

Three facts about right triangles:

(I) Area: R = 1/2 * a * b

(II) Perimeter: P = a + b + c

(III) Pythagorean Theorem:
a^2 + b^2 = c^2

The task is to find a, b, and c:

Step 1:

Of (I):

1/2 * a * b = R

a * b = 2 * R

Step 2:

Then from (II):

a + b + c = P

a + b = P – c

Step 3:

Squaring both sides:

(a + b)^2 = (P – c)^2

Step 4:

Expanding the left side:

a^2 + 2*a*b + b^2 = (P – c)^2

2*a*b + a^2 + b^2 = (P – c)^2

With a^2 + b^2 = c^2 (III) and a*b=2*R (Step 1):

2*(2*R) + c^2 = (P – c)^2

Simplify:

4*R + c^2 = P^2 – 2*P*c + c^2

4*R = P^2 – 2*P*c

c = (P^2 – 4*R)/(2*P)

Step 5:

Rearrange (I) to solve for a (we could solve for b
but the procedure is similar)

1/2 * a * b + R

a = (2*R)/b

Step 6:

Use (III) and Step 5 to solve for b:

a + b + c = P

a + b = P – c

(2*R)/b + b = P – c

2*R + b^2 = b*(P – c)

(at this point, c is known)

b^2 – b*(P – c) + 2*R = 0

Using the quadratic formula:

b = ( (P-c) ± √((P-c)^2 – 8*R) )/2

In conclusion:

c = (P^2 – 4*R)/(2*P)

b = ( (P-c) ± √((P-c)^2 – 8*R) )/2

a = (2*R)/b = P – a - c

Example:

Given R = 6 and P = 12

c = (12^2 – 4*6)/(2*12) = 5

b = ( (12-5) + √((12-5)^2 – 8*6) )/2 = 4

a = 12 – 5 – 4 = 3

c = (12^2 – 4*6)/(2*12) = 5

b = ( (12-5) - √((12-5)^2 – 8*6) )/2 = 3

a = 12 – 5 – 3 = 4

Solutions:

a = 3, b = 4, c = 5

or

a = 4, b = 3, c =5

Eddie

This blog is property of Edward Shore, 2016

try http://triancal.esy.es

ReplyDeleteVery informative article.Thank you author for posting this kind of article .

ReplyDeletehttp://www.wikitechy.com/view-article/write-a-c-program-to-find-area-of-triangle-with-example-and-explanation

Both are really good,

Cheers,

Venkat