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triangles and perfect squares

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triangles and perfect squares

by J N » Tue Mar 12, 2013 2:39 pm
Can someone provide an quick refresher on how these relate .


I saw a problem where the area was used to determine hypotenuse c or sides a and b using Pythagoras and perfect square formula.

a^2 + b^2 = c^2

area = 1/2 base x height = 1/2 a*b = 48

(a+b)^2 = a^2 + b^2 +2ab = c^2

=a^2 + b^2 +2(96)= c^2

.....
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by Anju@Gurome » Tue Mar 12, 2013 5:30 pm
J N wrote:Can someone provide an quick refresher on how these relate .


I saw a problem where the area was used to determine hypotenuse c or sides a and b using Pythagoras and perfect square formula.

a^2 + b^2 = c^2

area = 1/2 base x height = 1/2 a*b = 48

(a+b)^2 = a^2 + b^2 +2ab = c^2

=a^2 + b^2 +2(96)= c^2

.....
It would have been easier for me if you have posted the original problem and then asked your doubt. Anyway, the lines marked in blue and red are contradictory. In a right-angled triangle, if a and b are the lengths of the base and the leg of the triangle and c is the hypotenuse of the triangle, then (a² + b²) = c²

Also the area of the triangle = (base*height)/2 = ab/2

Now, from algebra (a + b)² = (a² + b² + 2ab) = (a² + b²) + 2ab = c² + 2ab
So, c² = (a + b)² - 2ab = (a + b)² - 4*(ab/2) = Square of the sum of leg and base + 4*(Area of the triangle)

Without the actual problem I cannot help you any further as only from area of the triangle, we cannot determine the hypotenuse. There must be some other information available.
Anju Agarwal
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