QUESTION 2
Easy
The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.
SOLUTION
1
Find the horizontal part
Differences of bases \(= 40 − 20 = 20\,cm\)
Each right triangle has base \(\frac{20}{2} = 10\,cm\)
Each right triangle has base \(\frac{20}{2} = 10\,cm\)
2
Find the height using Pythagoras Theorem
Hypotenuse, \(h\,=\, 26\,cm\)
Base, \(b\,= \,10\,cm\)
\(h = \sqrt{{26}^2 − {10}^2} = \sqrt{576} = 24\,cm\)
Base, \(b\,= \,10\,cm\)
\(h = \sqrt{{26}^2 − {10}^2} = \sqrt{576} = 24\,cm\)
3
Find the area
\(\begin{aligned} Area &= \frac{1}{2}(40 + 20) \,× \,24 \\ &= \frac{1}{2}\,×\,60\,×\,10\\ &= 720\,cm^2\end{aligned}\)
The area of the trapezium is 720 cm\(^2\).
The area of the trapezium is 720 cm\(^2\).
🏆
Final Answer : The area of the trapezium is 720 cm\(^2\).
Concept Note
The area of a trapezium is given by
\(Area = \frac{1}{2}(a + b)h\)
where,
\(a,b\) are the parallel sides
\(h\) is the perpendicular height