Question

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solution for question ​

Answer 1

Answer:

Step-by-step explanation:

From the picture we have Δ COD is similar to Δ AOB.

Since,

∠COD=∠AOB (opposite angle) , OBA∠CDO=∠OBA (transversal angle)

∠DCO=∠OAB (transversal angle).

Then

CD/AB=OD/OA

or 6/20 = OD/15

or, OD=4.5

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Answer 2

Given :

• In trapezium, ABCD,

1. Side AB $\parallel$ Side DC
2. Line DB is the transversal
3. Diagonals AC and BD intersects in point O.

• AB = 20
• DC = 6
• OB = 15

To Find :

• Length of OD.

Solution :

Side AB $\parallel$ side DC

line DB is the transversal.

➟ $\sf{\angle{CDB}}$ $\sf{\cong}$$\sf{\angle{ABD}}$ $\sf{\underbrace{Alternate\:angles\:theorem}}$

➟ $\sf{\angle{CDO}}$ $\sf{\cong}$$\sf{\angle{ABO\:\:\:(i)}}$

➟ In $\sf{\triangle{COD}}$ and $\sf{\triangle{AOB}}$

➟ $\sf{\angle{CDO}}$ $\cong$ $\sf{\angle{ABO}}$ [From (i)]

➟ $\angle{COD}$ $\cong$ $\angle{AOB}$ $\sf{\underbrace{Vertically\:opposite\:angles}}$

$\sf{\therefore{\triangle{COD}\sim{\triangle{AOB}}}}$$\sf{\underbrace{AA\:test\:of\:Similarlity}}$

$\sf{\therefore{\dfrac{CO}{AO}}}$ = $\sf{\dfrac{OD}{OB}}$ = $\sf{\dfrac{DC}{AB}}$ $\sf{\underbrace{C.S.S.T\:of\:congruent\:triangles}}$

According to the question, we need to calculate OD. Therefore, consider the pairs with the given data neglecting the other ratio of remaining sides.

➟ $\sf{\dfrac{OD}{15}}$ = $\sf{\dfrac{6}{20}}$

➟ $\sf{OD\:=\:{\dfrac{15\:\times\:6}{20}}}$

➟ $\sf{OD\:=\:{\dfrac{15\:\times\:3}{10}}}$

➟ $\sf{OD\:=\:15\:\times\:0.3}$

➟ $\sf{OD=4.5}$

$\sf{\therefore{\underline{Length\:of\:OD\:=\:4.5\:units}}}$