Question

in the given figure , chords ab and cd of the circle are produced to meet at o. prove that triangles odb and oca are similar . given that cd=2 cm and do= 6 and bo =3 . and also find the ratios of the areas of quadrilateral cabd and triangle cao​

Answer 1

Step-by-step explanation:

AB and CD are the chords of the circle. they are produced to meet at O.

CD = 2cm.

DO = 6cm.

BO = 3cm.

To prove:  

ODB ~ OCA

Proof:

In triangles ODB and OCA

<DOB = <COA

<ACD = <DBO [property of angle of cyclic quadrilateral]

ODB ~ OCA

Hence proved.

Answer 2

Answer:

Proof given below.

Ratio of areas = $\frac{55}{64}$

Step-by-step explanation:

Proof that ΔODB and ΔOAC are similar:

Clearly ABCD is a cyclic quadrilateral.

⇒ ∠ACD + ∠ABD = 180° (Opposite angles of a cyclic quadrilateral are supplementary)

Also, ∠ABD + ∠OBD = 180° (Linear pair)

From the above equations, we get ∠ACD = ∠OBD

Also ∠BOD = ∠COA (Common)

Therefore by AA similarity criterion, ΔODB and ΔOAC are similar.

Ratio of areas of quadrilateral CABD and ΔCAO:

We know that ratio of areas of two similar triangles is the square of the ratio of their corresponding side lengths.

⇒$\frac{ar(OBD)}{ar(OAC)}$ = $\frac{BO^2}{CO^2}$ = $\frac{BO^2}{(CD+DO)^2}$ = $\frac{3^2}{(2+6)^2}$ = $\frac{9}{64}$

So let ar(ΔOBD) = 9x and ar(ΔOCA) = 64x.

⇒ ar(CABD) = ar(ΔOCA) - ar(ΔOBD) = 64x - 9x = 55x

∴ $\frac{ar(CABD)}{ar(CAO)}$ = $\frac{55x}{64x}$ = $\frac{55}{64}$