The distance between parallel sides of a trapezium is 12 cm and the distance between mid-points of other sides is 18 cm. Find the area of the trapezium.
Answers
Answer:
Let ABCD is a trapezium in which AB | | CD and distance between AB and CD is
12 cms. Mid points of DA and BC are P and Q respectively. PQ = 18 cms.(given).
Thus , AB | | PQ | | CD. Join A to C , AC diagonal meets PQ at point R.
In triangle ACD , mid point of DA is P and PR | | CD therefore CD= 2.PR
In triangle CAB , mid point of BC is Q and QR | | AB therefore AB=2.QR
Area of trapezium ABCD =(1/2)×sum of parallel sides× distance between them.
= (1/2)×(AB+CD)×12. cm^2. , putting AB=2.QR and CD= 2.PR.
= (1/2)×(2.QR+2.PR)×12. cm^2.
= (QR+PR)×12.cm^2.
= ( PQ)×12. cm^2.
= 18×12. cm^2.
= 216. cm^2.
Answer:
The distance between the parallel sides of the trapezium being 12 cm and the distance between the mid points of the other two sides being 18 cm, the area of the trapezium is 12x18 = 216 sq cm
Step-by-step explanation:
1× sum of parallel sides = 18 cm
∴18×12= Area
Area = 216cm
I hope it will help You