# Finding the area of a trapezoid given vertices

I encountered this problem and did not know how to solve it:

I graphed it out - making a point for each vertice. I then decided there must be a 3 because when $$x=0$$ then $$y=3$$. But, according to that logic, it should be answer c - because when $$y=0$$ then $$x=3$$. But, that is not the answer.

Where did I go wrong, and how should I go about solving this problem when I am not given an answer key?

• I'm not sure what the problem is. Isn't the answer actually (c)? (Normally, we would write that $\int_{x=0}^3 (3+x)\,dx$ to make it clearer what the integrand is, but I don't think that's what the issue is here.) – Brian Tung Jul 22 at 3:39
• In the picture, didn't you click option d)? – Toby Mak Jul 22 at 3:42
• Yes, I did. But, why is the answer not answer d? – burt Jul 22 at 3:45
• Well, is $3-x$ equal to $6$ when $x = 3$? If not, then (d) is not correct. (Remember that $(3, 6)$ represents $x = 3, y = 6$.) – Brian Tung Jul 22 at 3:46
• But, when $y=0$ then $x$ is not equal to $-3$. Or, am I just getting majorly confused? – burt Jul 22 at 3:48

When you look at the picture of the trapezoid in question, to find its area you are actually integrating a line segment from $$(0,3)$$ to $$(3,6)$$ in $$dx$$.

The slope of this is line is $$(6-3)/(3-0) = 1$$ and plugging in the first point yields the equation $$y=x+3$$. Thus we will integrate $$(x+3)$$.

As for the limits, you need to go $$(0,3) \to (3,6)$$ in $$x$$, so $$x=0$$ to $$x=3$$...

• So I should find the two points that really include the area of the entire thing, find the equation of that line, and then take the integral of that? – burt Jul 22 at 3:54
• @burt that's the idea, when you integrate an area under some function... – gt6989b Jul 22 at 14:53

Well, the correct answer is c. The slope of the line that represents the upper part of the trapezoid should be $$m=\frac{3}{3}=1$$ (the run is $$3$$ and the rise is also $$3$$ because $$6-3=3$$) and the $$y$$-intercept of that line, $$b$$, is $$3$$. Plugging all that information into the slop-intercept form of a line $$y=mx+b$$ gives us:

$$y=x+3.$$

And to get the area under that curve, you would integrate it from $$0$$ to $$3$$:

$$\int_{0}^{3}(x+3)\,dx.$$

The equation of the segment joining $$(0,3)$$ to $$(3,6)$$ is $$y=3+x$$

You are finding the area under $$y=3+x$$ and over the $$x$$-axis from $$x=0$$ to $$x=3$$

The correct choice is $$(c)$$

• So I first find the equation of the line that the area under the line is the area of the trapezoid? – burt Jul 22 at 4:04
• Yes, of course because that is the one which you are going to integrate. – Mohammad Riazi-Kermani Jul 22 at 4:17

The area is $$\int_0^3 (x+3)dx = \frac{27}{2} = 13.5$$

This is the area between the curve $$y = x+3$$ and the $$x$$-axis, between $$x = 0$$ and $$x = 3$$. You are trying to calculate this area, right?