Find the largest possible area of a right angled triangle whose hypotenuse is 5cm long
Answers
Answer:
Answer is 6.25 square centimeters.
For a right angle triangle of sides a, b and hypotenuse c, the Pythagorean theorem states a-squared + b-squared = c-squared or a^2 + b^2 = c^2
The area of such triangle is (1/2)(base)(height) = (1/2)(a)(b).
So area is maximized when (1/2)ab is maximized
Now in this case, a^2 + b^2 = c^2 = 5^2= 25.
b^2 = 25 - a^2 so b = + sqrt(25-a^2)
Area= (1/2)(a)(sqrt[25-a^2]) = y. Note by inspection that area = minimum = 0 when a= 0 and when a =5
y is at maximum value when derivative y’ =0.
y’ = (1/2)(1)(sqrt[25-a^2]) + (1/2)(a)(1/2)(-2a)(sqrt[25-a^2])^(-1/2)
For simplicity, recall and substitute b = sqrt(25-a^2) and then reduce. So y’ = 0 when b - (a^2)(b)^(-1) = 0. Since b is not = 0 (why?), multiply both sides by b to get
y’ = 0 = b^2 - a^2 = (25 -a^2) - a^2 = 25 -2a^2, which = 0 when a^2 = 25/2 or a = +5/(sqrt2) or (5)(sqrt2)/2
Finally, using this value for a in the equation for Area yields: Area maximum = (1/2)(5/sqrt2)(sqrt[25/2]) = 25/4 or 6.25 square centimeters.
Note that in a 3–4–5 right triangle, the area is (1/2)(3)(4) = 6, which is a bit less than the maximum of 6.25 square centimeters.
Answer:
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