Question

In a trapezium ABCD, AB is parallel to DC, BC is perpendicular to DC and ∠BAD=45°. If DC=5cm , BC=4 cm, the area of the trapezium in sq cm is

Answer 1

Dropping a perpendicular \(DE\) onto \(AB\), the figure is divided into two parts - a rectangle of dimensions \(4×5\) and an isosceles triangle \(AED\).

Required answer = \(\text{Area of rectangle} + \text{Area of triangle}\) 

= \(4×5+\frac{1}{2}(4×4)=28\;cm^2\)

Answer 2

In \(\triangle EAD\): \(Tan \ 45\degree = \frac{DE}{AE}\)
⇒ \(1 = 4/AE = AE = 4\) cm Then\(AB = AE + EB\)
 ⇒ \(4+ 5\)
⇒ \(9 cm\) 
The area of trapezium = \((1/2) × (sum of parallel sides) × height \)
The area of trapezium = \((1/2) × (9 + 5) × 4\)
⇒\(2 × 14\)
⇒ \(28\ cm^2\) 
The area of the trapezium in sq cm is 28 cm2