# Triangle Area - Correct Procedure ???

• Oct 30th 2009, 03:55 PM
Monocerotis
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
• Oct 30th 2009, 04:42 PM
skeeter
Quote:

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 ...

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

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

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

$A_{max} = \frac{1}{2}(10.5)^2$
• Oct 30th 2009, 08:24 PM
Monocerotis
Quote:

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

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

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

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

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

lol yeah looks like I took the long road, thanks for checking it over I'll also write down your method for future problems.