enter the correct answer in the box.
the triangle shown is an isosceles triangle because the lengths of two legs are the same.
image of a right isosceles triangle
$3g^2 - 6g + 2$
what standard form polynomial expression represents the area of the triangle? hint: $\text{area} = \frac{1}{2}bh$.

Question

Accepted Answer

# Explanation:
## Step1: Identify base and height
The triangle is isosceles right - angled, so base $b = 3g^{2}-6g + 2$ and height $h=3g^{2}-6g + 2$ (since two legs are equal).
## Step2: Apply area formula
The formula for the area of a triangle is $A=\frac{1}{2}bh$. Substitute $b = 3g^{2}-6g + 2$ and $h = 3g^{2}-6g + 2$ into the formula:
$$
\begin{align*}
A&=\frac{1}{2}(3g^{2}-6g + 2)(3g^{2}-6g + 2)\
&=\frac{1}{2}(3g^{2}-6g + 2)^{2}
\end{align*}
$$
## Step3: Expand the square
First, expand $(3g^{2}-6g + 2)^{2}$ using the formula $(a + b + c)^{2}=a^{2}+b^{2}+c^{2}+2ab + 2ac+2bc$ (here $a = 3g^{2}$, $b=-6g$, $c = 2$):
$$
\begin{align*}
(3g^{2}-6g + 2)^{2}&=(3g^{2})^{2}+(-6g)^{2}+2^{2}+2\times(3g^{2})\times(-6g)+2\times(3g^{2})\times2+2\times(-6g)\times2\
&=9g^{4}+36g^{2}+4- 36g^{3}+12g^{2}-24g\
&=9g^{4}-36g^{3}+(36g^{2}+12g^{2})-24g + 4\
&=9g^{4}-36g^{3}+48g^{2}-24g + 4
\end{align*}
$$
## Step4: Multiply by $\frac{1}{2}$
Now multiply the expanded form by $\frac{1}{2}$:
$$
\begin{align*}
A&=\frac{1}{2}(9g^{4}-36g^{3}+48g^{2}-24g + 4)\
&=\frac{9}{2}g^{4}-18g^{3}+24g^{2}-12g + 2
\end{align*}
$$

# Answer: $\frac{9}{2}g^{4}-18g^{3}+24g^{2}-12g + 2$

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QUESTION IMAGE

Question

Explanation:

Step1: Identify base and height

Step2: Apply area formula

Step3: Expand the square

Step4: Multiply by \(\frac{1}{2}\)

Answer: