Mathematics, 23.09.2019 22:30 justijust500

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.

Find the area of the trapezoidal cross-section of the irrigation canal shown below. your answer will

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Answer from: amulets7017

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

Learn moreFind out the area of parallelogram The order of rotational symmetry of rhombus Find out the coordinates of the image of a point after the triangle is rotated 270° about the origin

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

Find the area of the trapezoidal cross-section of the irrigation canal shown below. your answer will

Answer from: Britny2386

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

The measurement of one of the parallel side is

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 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 $\left({w+\frac{h}{{\tan\theta }}}\right)h$ .

Learn more:

Learn more about the distance between two points on the number line Learn more about the distance between two coordinates of the line Learn more about what is the domain of the function on the graph? all real numbers all real numbers greater than or equal to 0 all real numbers greater than or equal to –2 all real numbers greater than or equal to –3

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 from: Quest

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.

Answer from: Quest

explanation:

step-by-step explanation:

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