Question

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

Answer 1

Side AB || Side DC and line DB is transversal

.'.
Side AB || side DC and line AC is transversal


.'. triangle DCO ≈ OAB ...AA test ... from (1) & (2)
≈ means Similar

DC/BA = DO/BO ... Ratio of corresponding sides of 2 similar triangles

DC/BA = DO/BO

6/20 = DO/15

DO × 20 = 6 × 15

.'. DO = 4.5

.'. OD = 4.5

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Answer 2

Answer:

OD=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