As we can see from the figure, DOB is a straight line.

Therefore, ∠DOC + ∠ COB = 180°

⇒ ∠DOC = 180° – 125° (Given, ∠ BOC = 125°)

= 55°

In ΔDOC, sum of the measures of the angles of a triangle is 180°

Therefore, ∠DCO + ∠ CDO + ∠ DOC = 180°

⇒ ∠DCO + 70° + 55° = 180°(Given, ∠ CDO = 70°)

⇒ ∠DCO = 55°

It is given that, ΔODC ∝ \( \dfrac{1}{4}\) ΔOBA,

Therefore, ΔODC ~ ΔOBA.

Hence, Corresponding angles are equal in similar triangles

∠OAB = ∠OCD

⇒ ∠ OAB = 55°

∠OAB = ∠OCD

⇒ ∠OAB = 55°

Answered by Abhisek | 1 year agoIn an trapezium ABCD, it is given that AB ∥ CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar(∆AOB) = 84 cm^{2}. Find ar(∆COD)

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