# Isosceles triangle of fixed perimeter P

• January 3rd 2010, 05:11 AM
dartanian24
Isosceles triangle of fixed perimeter P
Consider an isosceles triangle of fixed perimeter P.
a) If x equals the length of one of the two equal sides, express the area A as a function of x.

b) What is the domain of A?
• January 3rd 2010, 06:07 AM
abender
1)
$
A(x) = \frac{(P-2x)\sqrt{4Px-P^2}}{4}
$

(Note that A(x) above IS a function of x (and not x and P), since P is a fixed constant.)

2)
$
\text{Domain of } A: \frac{P}{4} < x < \frac{P}{2}
$

• January 3rd 2010, 06:11 AM
Wilmer
Quote:

Originally Posted by dartanian24
Consider an isosceles triangle of fixed perimeter P.
a) If x equals the length of one of the two equal sides, express the area A as a function of x.

To see where you're at:
can you calculate the height of this triangle in terms of P and x?
• January 3rd 2010, 06:17 AM
abender
Quote:

Originally Posted by Wilmer
To see where you're at:
can you calculate the height of this triangle in terms of P and x?

If he/she can calculate the height of the triangle in terms of P and x, that is solving it in terms of x, since x is the only variable in play; P is a fixed constant.

Good day!
-AB
• January 3rd 2010, 07:17 AM
Soroban
Hello, dartanian24!

Quote:

Consider an isosceles triangle of fixed perimeter $P$.

a) If $x$ equals the length of one of the two equal sides,
. . express the area $A$ as a function of $x\quad{\color{red}\hdots\;\text{ and }P}$

Code:

              A               *             /|\             / | \           /  |  \         x /  |  \ x         /    |h  \         /    |    \       /      |      \     B *-------*-------* C         ½b  D  ½b       : - - - b - - - :
Let $b$ = base of the triangle.
Let $h$ = height of the triangle.

Then: . $b + 2x \:=\:P \quad\Rightarrow\quad b \:=\:P-2x \;\;{\color{blue}[1]}\quad\Rightarrow\quad \frac{b}{2} \:=\:\frac{P - 2x}{2}\;\;{\color{blue}[2]}$

In right triangle $ADB\!:\;\;h^2 + \left(\frac{b}{2}\right)^2 \:=\:x^2
$

Substitute [2]: . $h^2 + \left(\frac{P-2x}{2}\right)^2 \:=\:x^2 \quad\Rightarrow\quad h^2 + \frac{P^2 - 4Px + 4x^2}{4} \:=\:x^2$

Multiply by 4: . $4h^2 + P^2 - 4Px + 4x^2 \:=\:4x^2 \quad\Rightarrow\quad h^2 \:=\:\frac{4Px-P^2}{4}$

Hence: . $h \:=\:\frac{\sqrt{4Px-P^2}}{2}\;\;{\color{blue}[3]}$

We know that: . $A \:=\:\tfrac{1}{2}bh$

Subnstitute [1] and [3]: . $A \;=\;\tfrac{1}{2}(P-2x)\frac{\sqrt{4Px-P^2}}{2}$

Therefore: . $A \;=\;\frac{1}{4}(P-2x)\sqrt{4Px-P^2}$

Quote:

b) What is the domain of $A$?
You probably mean "range".

The area has a minimum of 0 . . . when $b = 0.$

The area is a maximum when the triangle is equilateral.

. . Then: . $3x \:=\:P \quad\Rightarrow\quad x \:=\:\frac{P}{3}$

. . The area is: . $\frac{\sqrt{3}}{4}\left(\frac{P}{3}\right)^2 \;=\;\frac{\sqrt{3}}{36}P^2$

The range is: . $\left[0,\:\frac{\sqrt{3}}{36}P^2\right]$