Question

in a trapezium ABCD, AB is parallel to CD and AB2=CD. If
area of AOB=84cm, find the area of COD.​

Answer 1

Step-by-step explanation:

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

Answer 2

Answer:

ΔAOB~ΔCOD(AA similarity)

Step-by-step explanation:

Therefore corresponding sides are in proportion and by area theorem

(AB/CD)^2 = ar ΔAOB/arΔCOD

1/4=84/arΔCOD

84*4=arΔCOD

arΔCOD=336cm^2