Question

In a trapezium abcd ab parallel to cd and ab=2cd. If the area of triangle aob equal to 84cmsquare, find the area of triangle cod in easy method

Answer 1

In the triangles AOB and COD;

∠DOC = ∠ BOA [vertically opposite angles are equal]
∠ CDO = ∠ ABO

[ Since AB||CD with AC as transversal -alternate interior angles ]

∠ DCO = ∠ BAO 

thus Δ AOB ≈ Δ COD.

by the similarity rule, the ratio of the areas of the similar triangles is the ratio of the square of corresponding sides.


Area of AOB (AB)^2
__________=____

Area of COD (CD)^2

AB =2CD



84/A of COD= (2CD)^2 / (CD)^2

Area of COD = 84/4

= 21 sqcm

Answer 2

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