Area of an trapezoidal is express as :
[tex]Area\text{ = }\frac{Sum\text{ of Parallel sides}}{2}\times Height\text{ of trapezoid}[/tex]
So, to find the height of trapezium we draw a perpendicular from J to the base line LK and a perpendicular from I to the base line LK such that the MN= IJ
So, We have two right triangle JNK and IML
Apply Pythogaras in any one of the right triangle and find the length of perpendicular
In triangle JNK
[tex]\begin{gathered} \text{From Pythagoras ,} \\ \text{Hypotenuse }^2+Perpendicular^2_{}+Base^2 \\ JK^2=JN^2+NK^2 \\ \text{Substitute the valus and solve for JN,} \\ 5^2=JN^2+4^2 \\ JN^2=25-16 \\ JN^2=9 \\ JN=3 \\ \text{ So, The perpendicular of triangle JN is 3} \end{gathered}[/tex]
Since, the perpendicular JN is also the height of trapezium,
Substitute the values
Height JN =3
Parallel Sides IJ = 8 and LK = 16
So, the Area of Trapezium will be
[tex]\begin{gathered} \text{Area of trapezium IJKL =}\frac{Sum\text{ of parallel Sides}}{2}\times Height \\ \text{Area of trapezium IJKL =}\frac{IJ\text{ +LK}}{2}\times JN \\ \text{Area of trapezium IJKL =}\frac{8+16}{2}\times3 \\ \text{Area of trapezium IJKL =}\frac{24}{2}\times3 \\ \text{Area of trapezium IJKL =12}\times3 \\ \text{Area of trapezium IJKL =}36cm^2 \end{gathered}[/tex]
Answer : B) 36 cm²
