# Thread: function of area of isosceles triangle

1. ## function of area of isosceles triangle

The equal sides of an isosceles triangle are 2. If x is the base, express the area as a function of x.

The area of a triangle is 1/2 BH and the height of an Isosceles triangle I learned is :
H=square root of b^2-1/4a^2 (b is base, a is either of the equal sides)

So I got as an answer :

A = 1/2 X (x^2-1/4(2)^2)

Im pretty sure this is wrong. I keep trying to check it but Im getting it wrong. The official answer is:

A = 1/4X SQRT of 16-x^2

2. Since you know the three sides of the triangle $\displaystyle (2, 2, x)$ you should use Heron's Formula.

Heron's Formula: For any triangle of side lengths $\displaystyle a, b, c$ the semiperimeter is $\displaystyle s = \frac{a+b+c}{2}$ and the area is

$\displaystyle A = \sqrt{s(s-a)(s-b)(s-c)}$.

3. Here's another way to do it. If you draw the perpendicular from the vertex where the two equal sides meet to the opposite side, you get two congruent right triangles. The hypotenuse of each is one of the two equal sides and so has length 2. One of the legs is half the other side and so has length x/2. The other leg is the altitude of the original triangle and, taking its length to be "H", by the Pythagorean theorem, $H^2+ x^2/4= 4$. Then $H^2= 4- x^2/4= \frac{16- x^2}{4}$ so $H=\frac{\sqrt{16- x^2}}{2}$. That's exactly what you had!

Now, the area of a triangle is (1/2)*height*base which, here, is $A= \frac{1}{2}\frac{\sqrt{16- x^2}}{2}x= \frac{x\sqrt{16- x^2}}{4}$.

When you wrote "A = 1/2 X (x^2-1/4(2)^2)", first, it is not clear if "X" and "x" are the same or if you mean "X" as "multiplication". The crucial problem is that you have confused "a" and "b". Here, a= x while b= 2, not the other way around. That's the problem with just memorizing formulas. You have to be careful that you are assigning the correct values to the parameters.

4. Hello, frankinaround!

Here is yet another approach . . .

$\text{The equal sides of an isosceles triangle are 2.}$
$\text{If }x\text{ is the base, express the area as a function of }x.$
Code:
              *
/ \
/ @ \
/     \
2 /       \ 2
/         \
/           \
/             \
* - - - - - - - *
x

Let $\,\theta$ = the vertex angle.

Law of Cosines: . $\cos\theta \:=\:\dfrac{2^2+2^2 - x^2}{2(2)(2)} \;=\;\dfrac{8-x^2}{8}$

Using $\sin^2\!\theta + \cos^2\!\theta \:=\:1$, we find that: . $\sin\theta \:=\:\dfrac{x\sqrt{16-x^2}}{8}$

The area of a triangle is: . $A \;=\;\frac{1}{2}ab\sin C$
. .
One-half the product of two sides times the sine of the included angle.

Therefore: . $A \;=\;\frac{1}{2}(2)(2)\dfrac{x\sqrt{16-x^2}}{8} \;=\;\frac{1}{4}x\sqrt{16-x^2}$

5. Originally Posted by HallsofIvy
Here's another way to do it. If you draw the perpendicular from the vertex where the two equal sides meet to the opposite side, you get two congruent right triangles. The hypotenuse of each is one of the two equal sides and so has length 2. One of the legs is half the other side and so has length x/2. The other leg is the altitude of the original triangle and, taking its length to be "H", by the Pythagorean theorem, $H^2+ x^2/4= 4$. Then $H^2= 4- x^2/4= \frac{16- x^2}{4}$ so $H=\frac{\sqrt{16- x^2}}{2}$. That's exactly what you had!

Now, the area of a triangle is (1/2)*height*base which, here, is $A= \frac{1}{2}\frac{\sqrt{16- x^2}}{2}x= \frac{x\sqrt{16- x^2}}{4}$.

When you wrote "A = 1/2 X (x^2-1/4(2)^2)", first, it is not clear if "X" and "x" are the same or if you mean "X" as "multiplication". The crucial problem is that you have confused "a" and "b". Here, a= x while b= 2, not the other way around. That's the problem with just memorizing formulas. You have to be careful that you are assigning the correct values to the parameters.
Ohhhhhhhh. wow mistakes like that cost me hours. haha thanks man.