# How to find the maximum area of a triangle?

1. Nov 30, 2011

### justindizzle

1. The problem statement, all variables and given/known data

The hypotenuse of a right triangle has length of 10cm. Determine its maximum possible area.
c= 10cm

2. Relevant equations
a^2 + b^2 = c^2 ?
b*h*(1/2)

3. The attempt at a solution
just drew out the triangle so far, confused as to where to go from here
i assume ill be taking the derivative of some equation to find the max
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 30, 2011

### justindizzle

on second thought i don't think the pythagorean theorem would be relevant to the equation because the sides of the triangle are not changing and the derivative would imply that they do

3. Nov 30, 2011

### justindizzle

i think you can actually use both formulas
rearrange the pythagorean formula so that : b=10-a
(hypotenuse was 10) so a^2 + b^2 = 10^2
then substitute it into the base height formula :
= a(10-a) / 2
=10a - a^2 / 2
however when i solved for A and B i got 5*5/2 which = 12.5 but the answer is 25. what am i doing wrong ?

4. Nov 30, 2011

### Staff: Mentor

The sides have to change - the Pythagorean Theorem most definitely applies to this problem. The only thing that is constant is the hypotenuse.

5. Nov 30, 2011

### Staff: Mentor

???
You are assuming that sum of the two sides equals the hypotenuse. That's a very flat triangle.
One problem is that you have only 1 1/2 equations, not 2. Your area equation is
A = (1/2)ab

6. Nov 30, 2011

### justindizzle

okay so we know the triangle is a right triangle, and that the hypotenuse has a length of 10cm. We're trying to find the max area.

Pythagorean formula

a^2 + b^2 = c^2 (C = hypotenuse)

so according to what were given a^2 + b^2 = 10^
a^2 + b^2 = 100
we can rearrange this much into
b^2 = 100-a^2
and if you square root the whole equation it simplifies too
b=10-a correct?

i then substituted that into the base height formula which is (1/2)ab
=(1/2)a(10-a)
=(1/2)(10a-a^2)
=(5a)-(1/2)a^2 <--- multiplied the (1/2) in
=then i took the derivative to solve for A
=5-a
= a=5

Then putting "A" back into out previous formula b=10-a to solve for "b"
b=5

according to the b*h formula the area would be 5*5/2 = 12.5
but the answer is 25 in the back of the book ?

7. Nov 30, 2011

### Staff: Mentor

No. What you're saying is that, for example, $\sqrt{5^2 - 3^2} = 5 - 3 = 2$, which is not true.

8. Nov 30, 2011

### justindizzle

i'm hooped
how do i solve for b then ?
or can i square the entire equation into

a2*b2*(1/4)

9. Dec 1, 2011

### Staff: Mentor

You have b2 = 100 - a2. Then b = ±$\sqrt{100 - a^2}$. Your mistake earlier was in thinking you could simplify this further.
This is not an equation. An equation has an = in it somewhere.

10. Dec 1, 2011

### HallsofIvy

Staff Emeritus
What level course is this? I ask because there are several different ways to do this problem, the easier ones using "higher" level mathematics.

11. Dec 2, 2011

### PeterO

Draw a picture:

Did you know the angle in a semicircle is a right angle [common geometric proof]

Thus if the hypotenuse is the diameter of a circle, the right angle is also on the circle.

OK now have the hypotenuse as a horizontal base of a semicircle.

Area of a triangle is 1/2 x base x height

The base is the hypotenuse - and that is fixed in length.

What is maximum height of the triangle - remember the other vertex is on the semicircle.

Peter

12. Dec 2, 2011

### SammyS

Staff Emeritus
To make it an equation:
(Area)2 = (1/4)a2*b2

Since the area is positive, maximizing the square of the area will maximize the area.

Let x represent the square of length of one leg of the triangle, so that x = a2. The square of the length of the other leg is then given by b2 = (100-x) .

The area is then $\displaystyle (\text{Area})^2=\frac{1}{4}x(100-x)=-\frac{1}{4}x^2+25x\,.$

The square of the area is a quadratic function of the variable x. Can you find what value of x gives you an extremum for this function?