Question

in trapezium ABCD side ABllsideDC,diagonals AC and BD intersect in point O.If AB = 20,DC = 6,OB = 15 then find OD.

Answer 1

Answer:

Thus, the value of DO is 4.5.

Step-by-step explanation:

It is given that In trapezium ABCD side AB ll sideDC,diagonals AC and BD intersect in point O.

Now, since AB is parallel to DC, therefore ΔAOB is similar to ΔCOD

Thus, using the similarity conditions, we have

$\frac{DC}{BA}=\frac{DO}{BO}$

$\frac{6}{20}=\frac{DO}{15}$

$\frac{6{\times}15}{20}=DO$

$DO=\frac{9}{2}$

$DO=4.5$

Thus, the value of DO is 4.5.

Answer 2

Step-by-step explanation:

the solution is; OD =4.5