If AO - CO- OB - OD and BA - DC in the given figure then- find nbsp- the angle of AOB that will correspond to - C of COD is - nbsp-B-A-D-DOC

The correct option is B ∠AConsider nbsp; COD and nbsp; AOB AO = CO (Given) OB = OD ( Given ) BA = DC (Given) Hence nbsp;COD nbsp; ≅ nbsp; AOB by SSS ...

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Byju's Answer

Standard VII

Mathematics

SSS Congruence Criterion

If AO = CO, O...

Question

If AO = CO, OB = OD and BA = DC in the given figure
then, find the angle of $△ A O B$ that will correspond to $∠ C$ of $△ C O D$ is :

∠

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Solution

The correct option is D $∠$ A
Consider $△$ COD and $△$ AOB
AO = CO (Given)
OB = OD ( Given )
BA = DC (Given)
Hence $△$ COD $≅$ $△$ AOB
by SSS criterion for congruency of triangles.
When the triangles are equal then their corresponding parts are equal.

The angle that corresponds to $∠$ A in $△$ AOB is $∠$ C in $△$ COD.

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Similar questions

Q. In the given figure, rays OA, OB, OC and OD are such that
∠AOB = 56°, ∠BOC = 100°, ∠COD = x° and ∠DOA = 74°.
Find the value of x.

Q. Mark (✓) against the correct answer
In the given figure, rays OA, OB, OC and OD are such that ∠AOB = 50°, ∠BOC = 90°, ∠COD = 70° and ∠AOD = x°.
Then, the value of x is

(a) 50°
(b) 70°
(c) 150°
(d) 90°

Q. Question 6 (i)
In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar(DOC) = ar(AOB)

In the figure, rays OA, OB, OC, OD and OE have the common end point O. Show that $∠ A O B + ∠ B O C + ∠ C O D + ∠ D O E + ∠ E O A = 360^{\circ}$ .

In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:

(i) ar (DOC) = ar (AOB)

(ii) ar (DCB) = ar (ACB)

(iii) DA || CB or ABCD is a parallelogram.

[Hint: From D and B, draw perpendiculars to AC.]

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If AO = CO, OB = OD and BA = DC in the given figure then, find the angle of △AOB that will correspond to ∠C of △COD is :

If AO = CO, OB = OD and BA = DC in the given figure
then, find the angle of $△ A O B$ that will correspond to $∠ C$ of $△ C O D$ is :