Question

In the adjoining figure P Q R S is a rhombus, SQ and PR are the diagonals of the rhombus intersecting at point O. If angle OPQ=35 then find the value of angke ORS+ angle OQP​

Answer 1

The required sum = $90^o$

Step-by-step explanation:

We know that the diagonals of rhombus intersect at right angle.

• So, $\angle POQ = 90^o$ (refer the figure)

We can understand that;

• $\angle OPQ + \angle POQ + \angle OQP = 180^o$ (refer the figure)

• $35^o + 90^o + \angle OQP = 180^o$ (refer the figure)

• $\angle OQP = 180^o - 125^o = 55^o$ (refer the figure)

Since, in the rhombus, the opposite sides are parallel.

Therefore, PQ is parallel to RS and PR is transversal line. (refer the figure)

• So, $\angle OPQ = \angle ORS$ {they are internal alternate angles.}

• $\angle OPQ = \angle ORS = 35^o$

Hence, $\angle ORS + \angle OQP = 35^o + 55^o = 90^o$

The required sum = $90^o$

Answer 2

The answer is 90⁰

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