The area of the trapezium is [tex]\boxed{\left({w+\frac{h}{{\tan \theta }}}\right)h}[/tex] .
Further explanation:
The area is the trapezium can be calculated as,
[tex]{\text{area of trapaezium}}=\frac{1}{2}\times\left({{\text{sum of parallel sides}}}\right)\times{\text{height}}[/tex]
The trigonometry ratio used in the right angle triangles.
The tangent ratio can be written as,
[tex]\tan \theta = \frac{{{\text{length of the side opposite to }}\theta }}{{{\text{length of the side adjacent to }} \theta }}[/tex]
Here, base is the length of the side adjacent to angle [tex]\theta[/tex] and the length of side opposite to angle [tex]\theta[/tex] is perpendicular.
Step by step explanation:
Step 1:
Do naming the given trapezium as attached in the figure.
After naming we have [tex]PQRS[/tex] is a trapezium
Consider [tex]h[/tex] as the height of the trapezium in which [tex]{\text{X and W}}[/tex] are the points drawn on the line segment [tex]RS[/tex] .
The measurement of one of the parallel side [tex]PQ[/tex] is [tex]w[/tex]
It can be seen from the attached figure that [tex]RQ{\text{ and }}SP[/tex] are the transversal lines.
Therefore, by alternate interior angle property [tex]\angle QRW=\theta{\text{ and }}\angle PSX=\theta[/tex] .
The another parallel side can be written as [tex]SR=SX+XW+WR[/tex] .
Consider the sides as [tex]XW=w,SX=x,WR=x[/tex] .
Therefore, the parallel side can be written as [tex]SR=x+w+x[/tex] .
Step 2:
Now in the right angle triangle [tex]XSP[/tex] the tangent ratio can be applied as,
[tex]\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}[/tex]
Now substitute the value of [tex]x[/tex] in the equation [tex]SR=x+w+x[/tex] .
[tex]\begin{gathered}SR=\frac{h}{{\tan \theta }}+w+\frac{h}{{\tan \theta }}\hfill\\SR=2\frac{h}{{\tan \theta }}+w\hfill\\\end{gathered}[/tex]
Step 3:
The area of the trapezium can be calculated as,
[tex]\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}[/tex]
Further simplify the above equation.
[tex]\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}[/tex]
Therefore, the area of the trapezium is [tex]\left({w+\frac{h}{{\tan\theta }}}\right)h[/tex] .
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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.
