# Please i need help finding the cross sectional area trapezium!...

• Jan 21st 2013, 07:34 AM
ashicus
Please i need help finding the cross sectional area trapezium!...
1(b) A cutting is to be made through rough ground. The width of the bottom has to be a constant 4m and the sides of the cutting must be at a slope of 1 in 1. The depth s of the cutting at 6 m intervals are :-

(a) 1.8m (b) 1.4m (c) 1.4m (d) 1.8m (e) 2.0m (f) 2.6m
(g) 3.2m (h) 3.6m (i) 2.8m (j) 2.6m (k) 2.4m (l) 2.8m

(m) 3.8m

(ii) Determine the cross sectional area for each cross section.
• Jan 21st 2013, 08:35 AM
ibdutt
Re: Please i need help finding the cross sectional area trapezium!...
Use the expression for the area of a trapezoid, which is equal to 1/2 [ sum of the parallel sides x distance between them ]
• Jan 21st 2013, 03:31 PM
Soroban
Re: Please i need help finding the cross sectional area trapezium!...
Hello, ashicus!

Quote:

$\text{1. A channel is to be dug through level ground.}$
$\text{The width of the bottom has to be a constant 4m,}$
$\text{and the sides of the channel must be at a slope of 1-to-1.}$
$\text{The depth }h\text{ of the channel, at 6 m intervals, are:}$

. . $\begin{array}{c}\text{(a) 1.8m} \quad \text{(b) 1.4m} \quad \text{(c) 1.4m } \quad \text{(d) 1.8m} \quad \text{(e) 2.0m} \quad \text{(f) 2.6m } \quad \text{(g) 3.2m} \\ \text{(h) 3.6m } \quad \text{(i) 2.8m} \quad \text{(j) 2.6m} \quad \text{(k) 2.4m} \quad \text{(l) 2.8m} \quad \text{(m) 3.8m} \end{array}$

$\text{Determine the cross-sectional area for each cross section.}$

The cross section is an isosceles trapezoid.
Code:

      : - h - : - - 4 - - : - h - :       * * * * * * * * * * * * * * *         *    |          |    *           *  |h          |  *             * |          | * 45o               * * * * * * * - - -               : - - 4 - - :
The area of a trapezoid is: . $A \;=\;\tfrac{h}{2}(b_1+b_2)$
. . where $h$ is the height, and $b_1,\,b_2$ are the lengths of the two bases.

We have height $h$ and bases $4$ and $2h+4.$

Therefore: . $A \;=\;\tfrac{h}{2}(4+2h+4) \quad\Rightarrow\quad \boxed{A \;=\;h(h+4)}$

Now plug in the thirteen values for $h.$