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Bunuel
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Which of the figures below can be inscribed in a circle?

A. 1 only
B. 3 only
C. 1 & 3 only
D. 2 & 3 only
E. 1, 2 & 3

Useful property: a convex quadrilateral can be inscribed in a circle if and only its opposite angles are supplementary (supplementary angles are two angles that add up to 180°, whereas complementary angles are two angles that add up to 90°).

We can see on the diagram that only I and III meet that requirements.

Answer: C.

Hope it's clear.

I agree with "the opposite angles are supplementary" concept and I did the same.
Can this be a solution @Bunuel ?
1. for first diagram - Moreover, isn't the first one isoceles trapezium ? Isn't that sufficient ?
2. for third diagram - Parallelogram inscribed in a circle will become a rectangle (think of shape shifting)
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Bunuel
eybrj2
Which of the figures below can be inscribed in a circle?

A. 1 only
B. 3 only
C. 1 & 3 only
D. 2 & 3 only
E. 1, 2 & 3

Useful property: a convex quadrilateral can be inscribed in a circle if and only its opposite angles are supplementary (supplementary angles are two angles that add up to 180°, whereas complementary angles are two angles that add up to 90°). See Central Angle Theorem in Circles chapter of Math Book: math-circles-87957.html.

We can see on the diagram that only I and III meet that requirements.

Answer: C.

Hope it's clear.

Hi Bunuel,
By 'opposite angles' you mean diagonally opposite angles right? otherwise even II can be inscribed in a circle.
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Bunuel
eybrj2
Which of the figures below can be inscribed in a circle?

A. 1 only
B. 3 only
C. 1 & 3 only
D. 2 & 3 only
E. 1, 2 & 3

Useful property: a convex quadrilateral can be inscribed in a circle if and only its opposite angles are supplementary (supplementary angles are two angles that add up to 180°, whereas complementary angles are two angles that add up to 90°). See Central Angle Theorem in Circles chapter of Math Book: math-circles-87957.html.

We can see on the diagram that only I and III meet that requirements.

Answer: C.

Hope it's clear.

Hi Bunuel,
By 'opposite angles' you mean diagonally opposite angles right? otherwise even II can be inscribed in a circle.

Yes Diagonally opposite angles should sum up to 180 degree for the quadrilateral to be inscribed in a circle.

This property is a derived property from the one which says that angle subtended by any segment at the centre of the circle should be twice the angle subtended by the same segment on the circumference of the same circle.

Such Quadrilaterals which can be inscribed in a circle are called "Cyclic Quadrilateral"
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eybrj2
Which of the figures below can be inscribed in a circle?

(A) I only
(B) III only
(C) I & III only
(D) II & III only
(E) I, II & III

Attachment:
IMG_0414.jpg

------------------
If a quadrilateral can be inscribed in a circle then the sum of its opposite angles =180
only 2 of the quadrilaterals meet this criteria
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hi Bunuel

No arc of a circle can subtend an angle of >180 at the centre.
Is this a property that can be used in this sort of que?
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eybrj2
Which of the figures below can be inscribed in a circle?

(A) I only

(B) III only
(C) I & III only
(D) II & III only
(E) I, II & III

Attachment:
The attachment IMG_0414.jpg is no longer available

Bunuel, you never got your alternative solution...so here goes!!

In order for it to possible to inscribe a shape in a circle, we need to be able to find a point that is equidistant from the vertices. That point will be the center of the circle and the vertices will therefore all reside on the circumference. Lets look at our shapes.

I: If we draw a vertical line through the shape, all the points on the line are equidistant from the two upper vertices and also equidistant from the two lower vertices. If we pick the point on the line that is where the line meets the lower base (the red point), it will be closer to the to two lower vertices than it is to the two upper vertices. And if we pick the point on the line that is where the line meets the upper base (the green point), it will be closer to the two upper vertices than it is to the two lower vertices. That means that there must be some point between those two points where we are equidistant from the upper and lower vertices (the purple point). I works.

For II and III, I think we can simply visualize whether there's a way to draw a circle that hits all for vertices. If you get stuck and would benefit from a more thorough response, say the word and I'll supplement this post. For now, I'll go with II doesn't work and III does.

Answer choice C.
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