How to find the maximum area of a triangle?

In summary, the hypotenuse of a right triangle has a length of 10cm, and the maximum possible area is 25.
  • #1
justindizzle
5
0

Homework Statement



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

Homework Equations


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


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
 
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  • #2
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
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
justindizzle said:
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
The sides have to change - the Pythagorean Theorem most definitely applies to this problem. The only thing that is constant is the hypotenuse.
 
  • #5
justindizzle said:
i think you can actually use both formulas
rearrange the pythagorean formula so that : b=10-a
?
You are assuming that sum of the two sides equals the hypotenuse. That's a very flat triangle.
justindizzle said:
(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 ?

One problem is that you have only 1 1/2 equations, not 2. Your area equation is
A = (1/2)ab
 
  • #6
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
justindizzle said:
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?
No. What you're saying is that, for example, [itex]\sqrt{5^2 - 3^2} = 5 - 3 = 2[/itex], which is not true.
justindizzle said:
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 ?
 
  • #8
i'm hooped
how do i solve for b then ?
or can i square the entire equation into

a2*b2*(1/4)
 
  • #9
justindizzle said:
i'm hooped
how do i solve for b then ?
You have b2 = 100 - a2. Then b = ±[itex]\sqrt{100 - a^2}[/itex]. Your mistake earlier was in thinking you could simplify this further.
justindizzle said:
or can i square the entire equation into

a2*b2*(1/4)
This is not an equation. An equation has an = in it somewhere.
 
  • #10
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
justindizzle said:

Homework Statement



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

Homework Equations


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


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

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

a2*b2*(1/4)

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 [itex]\displaystyle (\text{Area})^2=\frac{1}{4}x(100-x)=-\frac{1}{4}x^2+25x\,.[/itex]

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?
 

Related to How to find the maximum area of a triangle?

1. What is the formula for finding the maximum area of a triangle?

The formula for finding the maximum area of a triangle is A = 1/2 * b * h, where b is the base of the triangle and h is the height. This formula is derived from the fact that the area of a triangle is equal to half the product of its base and height.

2. How do I determine the base and height of a triangle?

The base and height of a triangle can be determined by measuring the two sides that form a right angle, or by using the Pythagorean theorem to find the missing side. The side opposite the right angle is considered the height, and the other side is considered the base.

3. Can the maximum area of a triangle be found using any three sides?

No, the maximum area of a triangle can only be found using the base and height. In order to use the formula A = 1/2 * b * h, the base and height must be known. If only three sides are given, it may be possible to find the base and height using trigonometric functions, but it is not guaranteed.

4. Is the maximum area of a triangle affected by the shape of the triangle?

No, the maximum area of a triangle is not affected by the shape of the triangle. As long as the base and height are known, the formula A = 1/2 * b * h can be used to find the maximum area, regardless of the shape of the triangle.

5. How can I check if my calculated maximum area is correct?

You can check the accuracy of your calculated maximum area by using a different method to find the area, such as using Heron's formula or breaking the triangle into smaller right triangles and adding their areas together. If you get the same result, then your calculated maximum area is likely correct.

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