The Perimeter of a Isoceles triangle is 6cm. FIND THE MAXIMUM AREA!!!!!

[ClearCCTrue]

The Perimeter of a Isosceles triangle is 6cm. Find the lengths of t he sides of the triangle that maximum the area.

Heres what I tried so far, I drew a triangle with two lengths 'x' and one length 'y', where y is the base.


SO i got the Perimeter equation to be :
6 = 2x + y
y = 6 - 2x

(Since thats the entire base, the base length of one congruent triangle is (3 - x)

Then I proceeded to use pythagorean theorem in order to get the height of one of the congruent triangles

so, x^2 = (3-x)^2 + h^2

h^2 + 9 - 6x + = 0
thus I got h = sqrt(9-6x)

And I know, the area equation being:

A = xh/2

So I plugged the values in:

= x(sqrt(9-6x))/2

= sqrt(9x - 6x^2)/2

= (9x-6x^2)^1/2 (ALl divided by 2)

I Proceeded to derive and set it to zero, but I got a completely different answer from the actual which is '2'.
Any input would be very much appreciated!!!!

 

[Quaid]

The Perimeter of a Isosceles triangle is 6cm. Find the lengths of t he sides of the triangle that maximum the area.

Heres what I tried so far, I drew a triangle with two lengths 'x' and one length 'y', where y is the base.


SO i got the Perimeter equation to be :
6 = 2x + y
y = 6 - 2x

(Since thats the entire base, the base length of one congruent triangle is (3 - x)

Then I proceeded to use pythagorean theorem in order to get the height of one of the congruent triangles

so, x^2 = (3-x)^2 + h^2

h^2 + 9 - 6x + = 0
thus I got h = sqrt(9-6x)

You made a mistake, in solving for h. The expression highlighted in red above needs correcting.

And I know, the area equation being:

A = xh/2

This formula would be correct, if x were the base of your isosceles triangle. (x is not your base.)

The correct formula for you is A=yh/2

I got a completely different answer from the actual which is '2'.

Correct these two mistakes, and try again!

 

[ClearCCTrue]

You made a mistake, in solving for h. The expression highlighted in red above needs correcting.





This formula would be correct, if x were the base of your isosceles triangle. (x is not your base.)

The correct formula for you is A=yh/2





Correct your two mistakes, and try again!

My Guess is the mistake I made for solving for 'h' is the arithmetic sign? Is it so posed to be (-9 + 6x) ?

 

[Quaid]

My Guess is … 'posed to be -9 + 6x

Yes! But why guess?

h^2 + 9 - 6x = 0

The right-hand side is zero, but it's not zero after adding 6x and subtracting 9.

h^2 = 6x - 9

Only now do we take the positive square root.

= x(sqrt(9-6x))/2

= sqrt(9x - 6x^2)/2

Uh-oh. We cannot multiply x by what's inside a square root like that!

:?