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.
