Question

Ina trapezium ABCD, it is given that AB || CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar $\left(\Delta AOB\right)=84c{m}^2$. Find ar $\left(\Delta COD\right)$.

Answer 1

Consider triangle AOB and triangle COD

angle AOB = angle COD (Vertically opposite angles)

angle OAB = angle OCD (Alternate interior angles)

angle OBA = AngleODC (Alternate interior angles)

therefore,triangle AOB ~ triangle COD(AAA similarity criterion)

Area (AOB)=$84c{m}^2$(GIVEN)

AB=2 CD(GIVEN)

$\frac{area(AOB}{area\left(COD\right)}$ = $(\frac{AB}{CD}{)}^2$

[The ratio of areas of two similar triangles is the square of the ratio of the corresponding sides]

$\frac{area(AOB}{area\left(COD\right)}$= $(\frac{2CD}{CD}{)}^2$

$\frac{84}{area\left(COD\right)}$ = 4

Area (COD)= $\left(\frac{84}{4}\right)$

area (COD) = 21

Area(COD) = 21 $c{m}^2$