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A circle with centre O and diameter PR is inscribed in quad

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fdundo
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A circle with centre O and diameter PR is inscribed in quad [#permalink] New post   Updated on: 24 Sep 2019, 00:49
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A circle with center O and diameter PR is inscribed in quadrilateral ABCD. Find the value of x.

Show: :: OA
56°

Originally posted by fdundo on 23 Sep 2019, 22:02.
Last edited by Carcass on 24 Sep 2019, 00:49, edited 1 time in total.
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Re: A circle with centre O and diameter PR is inscribed in quad [#permalink] New post   21 Oct 2019, 11:08
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I am not sure about my approach but we know that if PR is diameter then PQR must be 90 degree so RPQ is 56. Since the RPQ and the x corresponds the same arc, they must be equal.
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Re: A circle with centre O and diameter PR is inscribed in quad [#permalink] New post   30 Jul 2020, 06:13
Can someone please explain the solution to this problem?
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Re: A circle with centre O and diameter PR is inscribed in quad [#permalink] New post   30 Jul 2020, 07:09
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fdundo wrote:
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geometry tough.jpeg


A circle with center O and diameter PR is inscribed in quadrilateral ABCD. Find the value of x.

Show: :: OA
56°


This question tests a circle property known as the Tangent-Chord Theorem (aka the Alternate Segment Theorem)

Image

This property is NOT tested on the GRE, so this question is out of scope



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Re: A circle with centre O and diameter PR is inscribed in quad [#permalink] New post   09 Jan 2021, 11:40
its alternate angles....
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Re: A circle with centre O and diameter PR is inscribed in quad [#permalink] New post   06 Feb 2023, 05:08
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A circle with centre O and diameter PR is inscribed in quadrilateral A [#permalink] New post   28 Jun 2023, 07:06
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A circle with centre O and diameter PR is inscribed in quadrilateral ABCD. Find the value of x?

Show: ::
56



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Re: A circle with centre O and diameter PR is inscribed in quadrilateral A [#permalink] New post   18 Jul 2023, 20:57
Can anyone explain this? I got angle p as 56 but how is it for x as well?
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A circle with centre O and diameter PR is inscribed in quadrilateral A [#permalink] New post   19 Jul 2023, 00:07
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In triangle OQR, OQ = OR. (Both OQ and OR are radii of given circle)
So, angle opposite to these two sides would also be equal.
So, we can say, in triangle OQR.
Angle OQR = Angle ORQ


Hence angle OQR = 34°
Now consider quadrilateral ABCD, Side AD of quadrilateral is tangent to the circle. We
know, the angle between the radius of a circle and tangent is 90°

So, we can say angle OQD = 90°
From the given figure, we can say angle OQD = angle OQR + angle RQD

Hence 90°= 𝑥° + 34°
Hence 𝑥°= 56°

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A circle with centre O and diameter PR is inscribed i [#permalink] New post   14 Sep 2023, 02:00
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A circle with centre O and diameter PR is inscribed in quadrilateral ABCD. Find the value of x?

Show: ::
56



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