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.
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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