Pls solve tia Share with your friends Share 0 Kunal Rao answered this Dear Student, Since F and E are the mid points of GH and GK, then FE∥HK Mid point theorem .....1In quad. FHKE we have, FE∥HK ⇒quad. FHKE is a trapezium.Since D and E are the mid points of HK and KG, thenDE∥HG Mid point theorem .....2In quad.FEDH, we haveFE∥HD From 1DE∥HF From 2So, FEDH is a parallelogram.We know that diagonal of a parallelogram divides it into 2 triangles of equal areas.So, ar∆FHD = ar∆DEF .......3Similarly, we get FDKE as a parallelogram.Then ar∆DEF= ar∆DEK .........4Similarly, we get FDEG is a parallelogram.Then ar∆DEF = ar∆EFG ........5Now, from 3, 4 and 5, we getar∆DEF= ar∆DEK = ar∆EFG = ar∆FHDNow, arFEKH = ar∆DEF+ ar∆DEK +ar∆FHDarFEKH = ar∆DEF + ar∆DEF + ar∆DEFarFEKH = 3 ar∆DEF= 3 ×14 ar∆GHK = 34ar∆GHK Regards 0 View Full Answer
