# Homework Help - Area of Trapezium

• Sep 3rd 2008, 12:12 PM
ajnokia
Homework Help - Area of Trapezium
Struggling with this question and so are my mates.

Attached is the Graph.

The line PQ has equation y=2x+4

(a) Find, without using calculus, the area of the shaded trapezium shown in the diagram.

(b) Express the area of this trapezium as a definite integral.

(c) Evaluate this integral

I just dont know where to go about tackling this question at all, I feel its the english putting me off. If anyone can help me to find the solutions to this question it will help me and assure I know what I am doing when I face similar questions in future.

Many Thanks

PS. The Q should be right at the end of the line with P..... the digit close to the P is a 0 not a Q, sorry about that
• Sep 3rd 2008, 12:22 PM
Chop Suey
The area of a trapezoid is:
$A_{\text{trapezoid}} = \frac{h (b_1 +b_2)}{2}$

Where h is the height and $b_1$ and $b_2$ are the upper and lower bases.

You might need to change your perspective. Remember that the bases have to be parallel sides!

Now for the next question....you have to know that the integral is the area under the curve (or line). Here, you're trying to find the area under the curve between 2 and 5.

As for question (c): I guess you have to integrate $f(x) = 2x + 4$
• Sep 3rd 2008, 12:27 PM
Chop Suey
Quote:

Originally Posted by Chop Suey
The area of a trapezoid is:
$A_{\text{trapezoid}} = \frac{h (b_1 +b_2)}{2}$

Where h is the height and $b_1$ and $b_2$ are the upper and lower bases.

You might need to change your perspective. Remember that the bases have to be parallel sides!

Or, you can think of the trapezium as a combination of rectangle and a triangle.

Recall that:
$A_{Triangle} = \frac{1}{2}bh$

$A_{Rectangle} = b \cdot h$
• Sep 3rd 2008, 01:03 PM
ajnokia
Quote:

Originally Posted by Chop Suey
Or, you can think of the trapezium as a combination of rectangle and a triangle.

Recall that:
$A_{Triangle} = \frac{1}{2}bh$

$A_{Rectangle} = b \cdot h$

So the base is 3. But I dont know the height for the rectangle or the triangle. So unless I'm missing something (which I probably am due to tiredness) I cant manage to work out the area of the triangle nor the rectangle?

I know what Im doing with and (b) and (c) now though (Happy) Thanks!
• Sep 3rd 2008, 01:09 PM
Chop Suey
The height of the rectangle is given by $f(x) = 2x+4$. The height of the triangle is given by $f(5) - f(2)$. Do you understand why?
• Sep 3rd 2008, 01:11 PM
ajnokia
Quote:

Originally Posted by Chop Suey
The height of the rectangle is given by $f(x) = 2x+4$. The height of the triangle is given by $f(5) - f(2)$. Do you understand why?

y = 2x+4 is for the line PQ so I dont understand how that becomes the height of the rectangle.
• Sep 3rd 2008, 01:19 PM
Chop Suey
The endpoints of the two segments lie on the line y = 2x + 4. To find the length of these two lines, we find f(5) and f(2).
• Sep 3rd 2008, 01:19 PM
leftyguitarjoe
Quote:

Originally Posted by ajnokia
y = 2x+4 is for the line PQ so I dont understand how that becomes the height of the rectangle.

your base goes from 2 to 5

just find y then x=5
• Sep 3rd 2008, 01:20 PM
leftyguitarjoe
Quote:

Originally Posted by Chop Suey
The endpoint of the two lines lie on the line y = 2x + 4. To find the length of these two lines, we find f(5) and f(2).

gah, ya beat me(Giggle)