What is the largest rectangle in terms of area that can be inscribed inside a right angled triangle with sides 5, 12 and 13cm, if one vertex of the rectangle is in contact with the triangle's hypotenuse?
Ron.
Ron.
A Triangle with a Rectangle within.
- Thread starter SydneyRoverP6B
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Well, if you put a rectange inside a right angled triangle so that the right angle forms part of the triangle, and the rectangle, and touches the hypoteneuse, then the resulting triangles (left over from the rectangle will have the same angles, and will therefore be 'similar' - i.e. Triangle ABC, ADE and EFC
If the rectangle is 4 high (80% of the height), the length will be 2.4 long, because the small triangle above the rectangle will be 1 x 2.4, a 1/5 replica of the 5 x 12.
So, the area of the rectangle can be shown as (h x a) x (l x 1-a) - l being the length, h is the height and a is between 0 and 1, being the scale. i.e. 0.5 is half the size.
A few basic calculations show that with a at 0.5 gives the biggest area, which is 15
Richard
If the rectangle is 4 high (80% of the height), the length will be 2.4 long, because the small triangle above the rectangle will be 1 x 2.4, a 1/5 replica of the 5 x 12.
So, the area of the rectangle can be shown as (h x a) x (l x 1-a) - l being the length, h is the height and a is between 0 and 1, being the scale. i.e. 0.5 is half the size.
A few basic calculations show that with a at 0.5 gives the biggest area, which is 15
Richard
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darth sidious
New Member
Phoenix said:
Oh flaming heck!
I thought Ron meant one side of the rectangle was on the hypotenuse! :-S That can be solved, but it's bloody hard!
i.e. like this!
darth sidious
New Member
After hitting my hard for a few minutes... ouch! 
:roll: :wink:
Taking the right angle side of the triangle to be at the origin...
Assuming that the '5' side is the vertical, and the '12' side is the horizontal.
We take the hypotenuse to be denoted by
x = 0, y = 5 and at x = 12, y = 0
m = (0 - 5)/(12 - 0) = - (5/12)
As y = 5 at x = 0, so y = 5(1 - x/12)
The area of the rectangle is given by xy, which is 5x (1 - x/12)
A = 5x (1 - x/12) = 5 (x - (x^2)/12)
dA/dx = 5(1 - x/6) = 0, so x = 6
d^2A/dx^2 = - 5/6, so the value at which dA/dx =0 yields the maximum value.
A = 30(1 - (1/2)) = 30/2 = 15
The maximum Area value of the internal rectangle is 15 (which, as I told Richard earlier, is half the entire triangle!)
(OK, I didn't really tell Richard, but I have to steal back some of my pride somehow or other!)
:roll: :wink:Taking the right angle side of the triangle to be at the origin...
Assuming that the '5' side is the vertical, and the '12' side is the horizontal.
We take the hypotenuse to be denoted by
x = 0, y = 5 and at x = 12, y = 0
m = (0 - 5)/(12 - 0) = - (5/12)
As y = 5 at x = 0, so y = 5(1 - x/12)
The area of the rectangle is given by xy, which is 5x (1 - x/12)
A = 5x (1 - x/12) = 5 (x - (x^2)/12)
dA/dx = 5(1 - x/6) = 0, so x = 6
d^2A/dx^2 = - 5/6, so the value at which dA/dx =0 yields the maximum value.
A = 30(1 - (1/2)) = 30/2 = 15
The maximum Area value of the internal rectangle is 15 (which, as I told Richard earlier, is half the entire triangle!
)(OK, I didn't really tell Richard, but I have to steal back some of my pride somehow or other!
)
Hello Richard,
Very well done and equally to Darth,..who I think was hitting his head,..well at least that is what I hope he was hitting..
The method that I used was the same as that used by Darth,..that is derive a function for the area of the rectangle, differentiate, equate to zero in order to obtain the critical point. Take the second derivative to test the concavity of the function and test the end points within the closed interval [0, 12]. The result is that x = 6 which delivers a maximum area of 15 square cm.
Good result all round...
Ron.
Very well done and equally to Darth,..who I think was hitting his head,..well at least that is what I hope he was hitting..
The method that I used was the same as that used by Darth,..that is derive a function for the area of the rectangle, differentiate, equate to zero in order to obtain the critical point. Take the second derivative to test the concavity of the function and test the end points within the closed interval [0, 12]. The result is that x = 6 which delivers a maximum area of 15 square cm.
Good result all round...
Ron.
darth sidious
New Member
yes, I meant hitting my head hard! :-S