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 cm2. Find ar(∆COD)
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