Question

Find the area of the trapezoidal cross-section of the irrigation canal shown below. Your answer will be in terms of h, w, and θ.

Answer 1

The area of the trapezium is $\boxed{\left({w+\frac{h}{{\tan \theta }}}\right)h}$ .

Further explanation:

The area is the trapezium can be calculated as,

  ${\text{area of trapaezium}}=\frac{1}{2}\times\left({{\text{sum of parallel sides}}}\right)\times{\text{height}}$

The trigonometry ratio used in the right angle triangles.

The tangent ratio can be written as,

$\tan \theta = \frac{{{\text{length of the side opposite to }}\theta }}{{{\text{length of the side adjacent to }} \theta }}$

Here, base is the length of the side adjacent to angle $\theta$  and the length of side opposite to angle $\theta$  is perpendicular.

Step by step explanation:

Step 1:

Do naming the given trapezium as attached in the figure.

After naming we have $PQRS$  is a trapezium  

Consider $h$  as the height of the trapezium in which ${\text{X and W}}$  are the points drawn on the line segment $RS$ .

The measurement of one of the parallel side $PQ$  is  $w$

It can be seen from the attached figure that $RQ{\text{ and }}SP$  are the transversal lines.

Therefore, by alternate interior angle property $\angle QRW=\theta{\text{ and }}\angle PSX=\theta$ .

The another parallel side can be written as $SR=SX+XW+WR$ .

Consider the sides as $XW=w,SX=x,WR=x$ .

Therefore, the parallel side can be written as $SR=x+w+x$ .

Step 2:

Now in the right angle triangle $XSP$  the tangent ratio can be applied as,

$\begin{aligned}\tan \theta &=\frac{{{\text{length of the side opposite to }}\theta }}{{{\text{length of the side adjacent to}}\ \theta }}\hfill\\\tan \theta &=\frac{h}{x}\hfill\\x&=\frac{h}{{\tan \theta }}\hfill\\\end{aligned}$

Now substitute the value of $x$  in the equation $SR=x+w+x$ .

$\begin{gathered}SR=\frac{h}{{\tan \theta }}+w+\frac{h}{{\tan \theta }}\hfill\\SR=2\frac{h}{{\tan \theta }}+w\hfill\\\end{gathered}$

Step 3:

The area of the trapezium can be calculated as,

$\begin{aligned}{\text{area of trapaezium}}&=\frac{1}{2}\times\left({{\text{sum of parallel sides}}}\right)\times{\text{height}}\\&=\frac{1}{2}\times\left({PQ+SR}\right)\times h\\&=\frac{1}{2}\times\left({w+w+\frac{{2h}}{{\tan \theta}}}\right)\times h\\&=\frac{1}{2}\times\left({2w+\frac{{2h}}{{\tan \theta }}}\right)\times h\\\end{aligned}$

Further simplify the above equation.

$\begin{gathered}{\text{area of trapaezium}}=\frac{1}{2}\times\left({2w+\frac{{2h}}{{\tan\theta}}}\right)\times h\hfill\\{\text{area of trapaezium}}=\left({w+\frac{h}{{\tan \theta }}}\right)h\hfill \\\end{gathered}$

Therefore, the area of the trapezium is $\left({w+\frac{h}{{\tan\theta }}}\right)h$ .

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Answer details:

Grade: High school

Subject: Mathematics

Chapter: Perimeters and area

Keywords: trapezium, area, parallel sides, sum, height, length, opposite side, adjacent side, trigonometry, tangent ratio, perpendicular, base, hypotenuse, right angle triangle.

Answer 2

[w + (h/tan θ)] × h

Further explanation

We aim is to find the area of the trapezoidal cross-section of the irrigation canal.

The formula for the area of the trapezoid is $\boxed{\boxed{ \ Area = \frac{1}{2} \times (a + b) \times h \ }}$

• a & b = parallel sides
• h = height

We assume the lower side is $\boxed{ \ a = w \ }$ and the upper side is b = x + w + x or $\boxed{ \ b = w + 2x \ }$. See the attached picture.

In the right triangles located on the left and right of the trapezoid, we calculate the value of x based on trigonometric ratios namely tan of theta.

$\boxed{ \ tan \ \theta = \frac{opposite}{adjacent} \ }$

$\boxed{ \ tan \ \theta = \frac{h}{x} \ }$

$\boxed{ \ x = \frac{h}{tan \ \theta} \ }$

Substitute the equation of x to $\boxed{ \ b = w + 2x \ }.$

We get $\boxed{ \ b = w + \frac{2h}{tan \ \theta} \ }$

Finally all components are complete and can be substituted into a formula to calculate the area of a trapezoid.

$\circ \ \boxed{a = w}\\ \circ \ \boxed{ \ b = w + \frac{2h}{tan \ \theta} \ }\\\circ \ and \ h$

$\boxed{ \ Area = \frac{1}{2} \times (w + w + \frac{2h}{tan \ \theta}) \times h \ }$

$\boxed{ \ Area = \frac{1}{2} \times (2w + \frac{2h}{tan \ \theta}) \times h \ }$

$\boxed{ \ Area = \frac{1}{2} \times 2(w + \frac{h}{tan \ \theta}) \times h \ }$

Hence, the area of the trapezoidal cross-section of the irrigation canal is $\boxed{\boxed{ \ Area = (w + \frac{h}{tan \ \theta}) \times h \ }}$

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Keywords: the trapezoidal, cross-section of the irrigation canal, the area, the answer will be in terms of h, w, and θ, trigonometric ratios, tan, opposite, adjacent, height, sides