Question

The parallel sides of a trapezium are 20cm and 10cm. it's nonparallel sides are both equal, each being 13 cm. find the area of the trapezium

Answer 1

ABCD be the given trapezium in which AB = 20 cm, DC = AE= 10 cm, BC = 13 cm and AD = 13cm. 


Through C, draw CE || AD, meeting AB at E. 

Draw CF ⊥ AB. 

Now, EB = (AB - AE) = (AB - DC) 

EB = (20- 10) cm = 10 cm; 

CE = AD = 13 cm; AE = DC = 13 cm. 

Now, in ∆EBC, we have CE = BC = 13 cm. 

It is an isosceles triangle. 

Also, CF ⊥ AB

So, F is the midpoint of EB. 

Therefore, EF = ¹/₂ × EB = 1/2× 10= 5cm. 

Thus, in right-angled ∆CFE, we have CE = 13 cm, EF = 5 cm. 

By Pythagoras’ theorem, we have 

CF = [√CE² - EF²] 

CF = √(13² - 5²) 

CF= √169-25= √144 = √12×12

CF= 12cm

Thus, the distance between the parallel sides is 12 cm. 

Area of trapezium ABCD = ¹/₂ × (sum of parallel sides) × (distance between them) 

Area of trapezium ABCD = ¹/₂ × (20 + 10) × 12 cm² 

Area of trapezium ABCD = 1/2×(30)× 12

Area of trapezium ABCD= 30×6 = 180 cm² 

Hence, Area of trapezium ABCD= 180 cm²