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

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In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O.

Practice Set 1.3 | Q 6 | Page 22
In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then Find OD.

Solution

Given:
side AB || side DC
AB = 20,
DC = 6,
OB = 15
In △COD and △AOB
∠COD = ∠AOB (Vertically opposite angles)
∠CDO= ∠ABO (Alternate angles, CD || BA and BD is a transversal line)
By AA test of similarity
△COD ∼ △AOB

∴ `(CD)/(AB) = (OD)/(OB)` (Corresponding sides are proportional)

\[ \Rightarrow \frac{6}{20} = \frac{OD}{15}\]

\[ \Rightarrow OD = 4 . 5\]

Answer:-

Given:

Trapezium ABCD with AB || DC
Diagonals AC and BD intersect at O
AB = 20, DC = 6, OB = 15

To find: OD

Proof:

In △COD and △AOB, we have:

∠COD = ∠AOB (vertically opposite angles)
∠CDO = ∠ABO (alternate angles, CD || BA and BD is a transversal)

Therefore, by AA test of similarity, △COD ~ △AOB.

By the properties of similar triangles, we have:

CD/AB = OD/OB (corresponding sides are proportional)

Substituting the given values, we get:

6/20 = OD/15

Simplifying, we get:

OD = 4.5

Therefore, OD is equal to 4.5 units.

Chapter 1. Similarity- Practice Set 1.3 – Page 22

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