In a trapezium ABCD ABl|CD .EF is drawn parallel to CD such that AE:DE=2:3 .If AB=10cm and CD=15cm then the length of EF will be (1) 6cm (2) 8cm (3) 10cm (4) 12cm
Answers
Given : In a trapezium ABCD ABl|CD .EF is drawn parallel to CD such that AE:DE=2:3 . AB=10cm and CD=15cm
To Find : length of EF
Solution:
Join BD intersecting EF at O
EO ║ AB ( as EF ║ CD & AB || CD )
in Δ ABD
EO ║ BD
Hence DE /AE = DO/BO => DO/BO = 3/2
& DE/AD = EO/AB
=> DE/(DE + AE) = EO/AB
=> 3/5 = EO/AB
=> EO = 3AB/5 = 3(10)/5 = 6 cm
now in Δ CDB
OF ║ CD
DO/BO = CF/BF = 3/2
=> BF/CF = 2/3
=> BC/CF = 5/3
=> BC/BF = 5/2
BC/BF = CD/FO
=> 5/2 = CD/FO
=> FO = 2CD/5 = 2(15)/5 = 6 cm
EO + FO = EF = 6 + 6 = 12 cm
Length of EF will be 12 cm
Learn More:
trapezium ABCD in which ab is parallel to CD are 2 X + 50 and 2x ...
https://brainly.in/question/11100485
in a Trapezium ABCD ab parallel DC a b equal to 30 cm BC equal to ...
https://brainly.in/question/13886159
Answer:
Hope the solution helps you!
