Question

in a trapezium abcd, ab parallel to cd and ab=2cd. if the area of triangle aob=84cm2 find area of triangle cod

Answer 1

HELLO DEAR,

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