Question

In trapezium ABCD, (Figure 1.60)side AB || side DC, diagonals AC andBD intersect in point O. If AB=20,DC=6,OB=15 then find OD.

Answer 1

hope you understand my answer

Answer 2

Answer:

OD = 4.5

Step-by-step explanation:

In the given figure,

AB || CD is given to us,

AB = 20

DC = 6

OB = 15

Now,

In triangle COD and triangle AOB,

∠DCO = ∠OAB (alternate interior angles)

∠CDO = ∠OBA (alternate interior angles)

and,

∠COD = ∠AOB (vertically opposite angles)

So,

ΔCOD ≈ ΔAOB (By, AAA similarity)

Now,

From CPCT we can say that,

$\frac{OD}{OB}=\frac{CD}{AB}\\So,\\\frac{OD}{15}=\frac{6}{20}\\So,\\OD=6\times \frac{15}{20}\\OD=\frac{9}{2}\\OD=4.5$

Therefore, the length of the side OD is given by,

OD = 4.5