# Thread: Triangle Area - Correct Procedure ???

1. ## Triangle Area - Correct Procedure ???

The sum of the base and the height of a triangle is 21cm. Determine the maximum area of the triangle, in square centimeters.
Solution
Area of a triangle:½ (b)(h)
b: base measurement of the triangle, h: height measurement of the triangle.
b + h = 21 || b = 21 – h

21h – h^2 = 0

h^2 – 21h = 0

h^2 – 21h +110.25 – 110.25 = 0

h^2 – 21h +110.25 = 110.25

(h-10.5)^2 = 110.25

√(h-10.5)^2 = √110.25

h-10.5 = ± 10.5

h = 10.5 ± 10.5 , h = 10.5 + 10.5, h = 21 || h = 10.5 – 10.5, h = 0

h = 10.5 + 10.5, h = 21

Divide 21 by 2

b = 21 – (h), b = 21 – (10.5), b= 10.5

h = 10.5, b = 10.5

Now plug in these values to find the area of the triangle.
Therefore, the area of the triangle, is 55.125 cm2

2. Originally Posted by Monocerotis
The sum of the base and the height of a triangle is 21cm. Determine the maximum area of the triangle, in square centimeters.
Solution
Area of a triangle:½ (b)(h)
b: base measurement of the triangle, h: height measurement of the triangle.
b + h = 21 || b = 21 – h
you sure took the long road to get there ...

$\displaystyle A = \frac{1}{2}(21-h)h$

$\displaystyle A = \frac{1}{2}(21h - h^2)$

max value at $\displaystyle h = \frac{-b}{2a} = \frac{-21}{-2} = 10.5$

$\displaystyle A_{max} = \frac{1}{2}(10.5)^2$

3. Originally Posted by skeeter
you sure took the long road to get there ...

$\displaystyle A = \frac{1}{2}(21-h)h$

$\displaystyle A = \frac{1}{2}(21h - h^2)$

max value at $\displaystyle h = \frac{-b}{2a} = \frac{-21}{-2} = 10.5$

$\displaystyle A_{max} = \frac{1}{2}(10.5)^2$
lol yeah looks like I took the long road, thanks for checking it over I'll also write down your method for future problems.