Finding the Minimum Hypotenuse of a Right Angled Triangle | Hypotenuse Homework

[cpashok]

Homework Statement

a and b are two perpendiculars drawn to the sides from a point on the hypotenuse of a right angled, isoceles triangle.

Find the minimum value of the hypotenuse in terms of a and b.

Homework Equations

ABC is a right angled, isoceles triangle, with AC, the hypotenuse, h.

X is a point on AC, from which the perpendiculars a and b are drawn to the sides AB and BC respectively.

AB = BC= h / (sq.root 2)

The Attempt at a Solution

Tried to represent h in terms of a single variable (some angle theta and trigonometric ratios of theta). This would eliminate two variables a and b. With this, h can be differentiated and equated to zero for Minima, and confirming it with 2nd derivative. But could not manage eliminating a and b and converting to a single variable equation.

 

[Mark44]

Homework Statement

a and b are two perpendiculars drawn to the sides from a point on the hypotenuse of a right angled, isoceles triangle.

Find the minimum value of the hypotenuse in terms of a and b.

Homework Equations

ABC is a right angled, isoceles triangle, with AC, the hypotenuse, h.

X is a point on AC, from which the perpendiculars a and b are drawn to the sides AB and BC respectively.

AB = BC= h / (sq.root 2)

The Attempt at a Solution

Tried to represent h in terms of a single variable (some angle theta and trigonometric ratios of theta). This would eliminate two variables a and b. With this, h can be differentiated and equated to zero for Minima, and confirming it with 2nd derivative. But could not manage eliminating a and b and converting to a single variable equation.

Since this is an isosceles right triangle, theta = pi/4, which it doesn't seem you have used. I don't know whether this is a helpful clue.

 

[cpashok]

Since this is an isosceles right triangle, theta = pi/4, which it doesn't seem you have used. I don't know whether this is a helpful clue.

Hi! Thanks for your time. I was not referring to theta as angle BAC or BCA, which is pi/4. For a point X, a and b are fixed, and are related (except that I am unable to figure it out straight). If I can represent h as an equation of a alone, or b alone, or some other variable theta on which a and b are dependent, then I can differentiate h and equate to 0 for Minima. This is where I am stuck. Or, is there some other way?

 

[cpashok]

Why is the attachment pending approval?
why is my question being unanswered?
Believe me i tried my best before posting up the question here.
Somebody please help me?

 

[gabbagabbahey]

Why is the attachment pending approval?

All attachments are screened by site admin before they are approved to be hosted at PF. This sometimes takes days, so it is quicker for you to host the image file at a free hosting site like imageshack.us and simply post a link to the image.

why is my question being unanswered?

Until we can view your attached file, we have no way of knowing where your mistakes might be.

 

[cpashok]

http://img218.imageshack.us/my.php?image=mathsrv7.gif

 

[gabbagabbahey]

Homework Statement

a and b are two perpendiculars drawn to the sides from a point on the hypotenuse of a right angled, isoceles triangle.

Find the minimum value of the hypotenuse in terms of a and b.

This question doesn't make sense to me; the length of the hypotenuse doesn't actually depend on 'a' and 'b', so it can't be minimized with respect to those variables.

Are you sure the question isn't "Find the minimum length of the line segment AX in terms of a and b"?

Is there more to the question than what you have posted here?

 

[cpashok]

This question doesn't make sense to me; the length of the hypotenuse doesn't actually depend on 'a' and 'b', so it can't be minimized with respect to those variables.

Yes you're right. I went to school today and got the question sorted out, it wasn't isosceles. A friend of mine verbally stated out the question he had picked up from some guide book, and I seem to have taken it down wrongly, mixed up multiple questions!

Sorry about that. I panicked as I had an exam today!

The correct question is:

A point on the hypotenuse is at a distance a and b from the sides of the triangle. show that the maximum length of hypotenuse is (a^(2/3) +b^(2/3))^(3/2).

btw, i had this question on the exam and i solved it correctly.

I sincerely apologize for this mistake once again.
Thanks for your help.