How to find the maximum area of a triangle?

[justindizzle]

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

 

[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

 

[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 ?

 

[Mark44]

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.

 

[Mark44]

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.

(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

 

[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 ?

 

[Mark44]

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, \sqrt{5^2 - 3^2} = 5 - 3 = 2, which is not true.

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 ?

 

[justindizzle]

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

a²*b²*(1/4)

 

[Mark44]

i'm hooped
how do i solve for b then ?

You have b² = 100 - a². Then b = ±\sqrt{100 - a^2}. Your mistake earlier was in thinking you could simplify this further.

or can i square the entire equation into

a²*b²*(1/4)

This is not an equation. An equation has an = in it somewhere.

 

[HallsofIvy]

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

 

[PeterO]

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

 

[SammyS]

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 \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?