#### Question

ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that: (i) ΔAOB and ΔCOD (ii) If OA = 6 cm, OC = 8 cm,

Find:(a) `("area"(triangleAOB))/("area"(triangleCOD))`

(b) `("area"(triangleAOD))/("area"(triangleCOD))`

#### Solution

We have,

AB || DC

In ΔAOB and ΔCOD

∠AOB = ∠COD [Vertically opposite angles]

∠OAB = ∠OCD [Alternate interior angles]

Then, ΔAOB ~ ΔCOD [By AA similarity]

(a) By area of similar triangle theorem

`("area"(triangleAOB))/("area"(triangleCOD))="OA"^2/"OC"^2=6^2/8^2=36/64=9/16`

(b) Draw DP ⊥ AC

`therefore("area"(triangleAOD))/("area"(triangleCOD))=(1/2xxAOxxDP)/(1/2xxCOxxDP)`

`="AO"/"CO"=6/8=3/4`

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#### APPEARS IN

Solution Abcd is a Trapezium in Which Ab || Cd. the Diagonals Ac and Bd Intersect at O. Prove That: (I) Triangle Aob and Triangle Cod (Ii) If Oa = 6 Cm, Oc = 8 Cm, Find:(A) `("Area"(Triangleaob))/("Area"(Trianglecod))` (B) `("Area"(Triangleaod))/("Area"(Trianglecod))` Concept: Areas of Similar Triangles.