Question

In a trapezium ABCD, AB || CD and AB = 2CD. If the diagonals AC and B
area of AOB is
(A) area (COD)
(B) 2area (COD)
(C) 4area (COD)
(D) 3area (COD)​

Answer 1

IN ∆ AOB and COD

< AOB= < COD (Vertically opposite angles)

< OAB= < OCD(Alternate interior angles)

< OBA = <ODC(Alternate interior angles)

therefore,∆ AOB ≈ ∆ COD(by AAA similarity criterion)

Area (AOB) = 84cm²(GIVEN)

AB =2CD(GIVEN)

area(AOB)/area(COD)=(AB/CD)²[The ratio of areas of two similar triangles is the square of the ratio of the corresponding sides]

area (AOB)/area(COD)=(2CD/CD)²

84/area (COD)=(2)²

area (COD)/84=1/4

area (COD)=1/4*84

area(COD)=21cm²

I HOPE ITS HELP YOU DEAR,

THANKS