The parallel sides of an isosceles trapezium are 48 centimetre and 24 centimetre each of its non parallel side is 26 metre long find the area of trapezium
Answers
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