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Find the area of a triangle with a base of [tex]\(3x^2y^2\)[/tex] and a height of [tex]\(4x^4y^3\)[/tex].

Sagot :

To find the area of a triangle, we use the formula:

[tex]\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]

Given that the base of the triangle is [tex]\( 3x^2y^2 \)[/tex] and the height is [tex]\( 4x^4y^3 \)[/tex], we can substitute these values into the formula.

First, let's plug in the values for the base and height:

[tex]\[ \text{Area} = \frac{1}{2} \times (3x^2y^2) \times (4x^4y^3) \][/tex]

Next, we perform the multiplication inside the parentheses. When multiplying these terms, we multiply the coefficients (numerical parts) and add the exponents of like variables:

[tex]\[ (3x^2y^2)(4x^4y^3) = 3 \times 4 \times x^{2+4} \times y^{2+3} \][/tex]

[tex]\[ = 12x^6y^5 \][/tex]

So, our expression now looks like this:

[tex]\[ \text{Area} = \frac{1}{2} \times 12x^6y^5 \][/tex]

Finally, we multiply by [tex]\(\frac{1}{2}\)[/tex]:

[tex]\[ \frac{1}{2} \times 12x^6y^5 = 6x^6y^5 \][/tex]

Therefore, the area of the triangle is:

[tex]\[ \boxed{6x^6y^5} \][/tex]
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