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# In the figure shown, quadrilateral ABCD is inscribed in a circle of

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In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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03 Jul 2016, 09:51
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72% (01:13) correct 28% (01:18) wrong based on 829 sessions

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In the figure shown, quadrilateral ABCD is inscribed in a circle of radius 5. What is the perimeter of quadrilateral ABCD?

(1) The length of AB is 6 and the length of CD is 8.
(2) AC is a diameter of the circle.

Attachment:

2016-07-03_2149.png [ 15.81 KiB | Viewed 27992 times ]

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In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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03 Jul 2016, 10:11
7
Perimeter of ABCD = AB + BC + CD + AD = ?

St1: The length of AB is 6 and the length of CD is 8. --> AB = 6 and CD = 8
Not Sufficient as value of BC and AD is not known.

St2: AC = diameter = 10 --> ABC and ADC are right angled triangles. Values of the sides of the quadrilateral cannot be derived.
Insufficient

Combining St1 and St2: AB = 6, CD = 8, ABC and ADC are right angled triangles
In ABC, AC = 10, AB = 6 --> BC = sqrt(10^2 - 6^2) = 8
In ADC, AC = 10, CD = 8 --> AD = sqrt(10^2 - 8^2) = 6
Perimeter = AB + BC + CD + AD = 28
Sufficient

##### General Discussion
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In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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Updated on: 04 Jul 2016, 19:25
we need 2 imp info-

1)ac is diamter
2)ac is perpendicular to BD

combining both we don't know AC is perpndclr to BD

SO C

we can find the lengths from pythagorus theorem.

Originally posted by hsbinfy on 03 Jul 2016, 22:29.
Last edited by hsbinfy on 04 Jul 2016, 19:25, edited 1 time in total.
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In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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03 Jul 2016, 22:40
1
Vyshak wrote:
Perimeter of ABCD = AB + BC + CD + AD = ?

St1: The length of AB is 6 and the length of CD is 8. --> AB = 6 and CD = 8
Not Sufficient as value of BC and AD is not known.

St2: AC = diameter --> We can infer that AC bisects line segment BD
Let O be the point of intersection of AC and BD.
Triangles AOD and AOB are congruent --> Therefore AB = AD
Triangles COB and COD are congruent --> Therefore BC = CD

AC = 10
Not Sufficient to determine the perimeter

Combining St1 and St2: Perimeter = AB + BC + CD + AD = 2(AB + CD) = 2(6 + 8) = 28
Sufficient

St2: AC = diameter --> We can infer that AC bisects line segment BD

how can you infer above statement until it is given AC is perpendicular to BD?

it will only bisect if AC is perpendicular to BD OR AC and BD intersect at centre of circle
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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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03 Jul 2016, 23:33
1
hsbinfy wrote:
Vyshak wrote:
Perimeter of ABCD = AB + BC + CD + AD = ?

St1: The length of AB is 6 and the length of CD is 8. --> AB = 6 and CD = 8
Not Sufficient as value of BC and AD is not known.

St2: AC = diameter --> We can infer that AC bisects line segment BD
Let O be the point of intersection of AC and BD.
Triangles AOD and AOB are congruent --> Therefore AB = AD
Triangles COB and COD are congruent --> Therefore BC = CD

AC = 10
Not Sufficient to determine the perimeter

Combining St1 and St2: Perimeter = AB + BC + CD + AD = 2(AB + CD) = 2(6 + 8) = 28
Sufficient

St2: AC = diameter --> We can infer that AC bisects line segment BD

how can you infer above statement until it is given AC is perpendicular to BD?

it will only bisect if AC is perpendicular to BD OR AC and BD intersect at centre of circle

Yes we cannot infer from Statement 2 that the two triangles are congruent. Thanks for spotting it.

But the answer is still C.

Since AC is the diameter, ABC and ADC are right angled.

In ABC, AC = 10, AB = 6 --> BC = sqrt(10^2 - 6^2) = 8
In ADC, AC = 10, CD = 8 --> AD = sqrt(10^2 - 8^2) = 6

Perimeter of the triangle = 6 + 8 + 6 + 8 = 28.
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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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10 Jul 2016, 06:36
2
Given: Radius of Circle = 5
Perimeter of ABCD AB + BC + CD + AD = ?

Statement 1: The length of AB is 6 and the length of CD is 8. --> AB = 6 and CD = 8
Not Sufficient, since the value of BC and AD is not known.

Statement 2: AC = diameter = 10 --> ABC and ADC are right angled triangles. Values of the sides of the quadrilateral cannot be derived.
Insufficient

Combining St1 and St2: AB = 6, CD = 8, ABC and ADC are right angled triangles
Perimeter can be derived.

Sample Calculation:
ABC right angle triangle, AC =10, AB = 6 $$BC^2 = AC^2 - AB^2$$
BC = 8

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In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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10 Jul 2016, 18:04
3
Top Contributor
1

In the figure shown, quadrilateral ABCD is inscribed in a circle of radius 5. What is the perimeter of quadrilateral ABCD?

(1) The length of AB is 6 and the length of CD is 8......>no information about the lengths of other two sides(information required because we don't know which specific quadrilateral is quadrilateral ABCD above),Insufficient

(2) AC is a diameter of the circle.........>So now we know AC=10 and its opposite angle $$\angle$$ABC=$$90^{\circ}$$(for $$\triangle$$ ABC) and $$\angle$$ADC=$$90^{\circ}$$(for $$\triangle$$ ADC),But still we don know the lengths of any sides of the Quadrilateral ABCD,Insufficient

(1)+(2) AB=6,AC=10 and $$\angle$$ABC=$$90^{\circ}$$ so According to pythagorean triple BC=$$\sqrt{AC^2+AB^2}$$=$$\sqrt{10^2-6^2}$$=8

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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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27 Apr 2017, 14:40
each alone is not sufficient.
1+2 => we have 2 right triangles (properties of inscribed triangles in a circle, where the hypotenuse is the diameter).
we know 2 sides, we can find 3rd side for each triangle.
in the end, we can find the perimeter.
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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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29 Sep 2017, 04:47
Hi,

And one of the properties of a cyclic quadrilateral is AB + CD = AD + BC (based on given fig.)

So shouldn't A be the answer choice?

Can someone explain pl.

Thanks
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Posts: 52431
Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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29 Sep 2017, 05:14
1
Ashokshiva wrote:
Hi,

And one of the properties of a cyclic quadrilateral is AB + CD = AD + BC (based on given fig.)

So shouldn't A be the answer choice?

Can someone explain pl.

Thanks

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This property is not true.
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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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06 Dec 2017, 04:21
(1) is very easily insuff but...

If we know that AC is the diameter of the circle and thus = 10, why wouldn't we say suff for (2) just off of the Pythagorean triplet 6-8-10?

That would tell me that AB/AD = 6 and BC/DC = 8.

6+6+8+8 = 28
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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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06 Dec 2017, 22:38
1
Roosterbooster wrote:
(1) is very easily insuff but...

If we know that AC is the diameter of the circle and thus = 10, why wouldn't we say suff for (2) just off of the Pythagorean triplet 6-8-10?

That would tell me that AB/AD = 6 and BC/DC = 8.

6+6+8+8 = 28

Hi

Based on statement 2, yes we know that both ABC and ADC are right angle triangles with hypotenuse AC=10. But we cannot just assume that a right angle triangle with hypotenuse 10 will only be a 6-8-10 triangle. It could also be 5√2-5√2-10 triangle OR it could be 2√5-4√5-10 triangle.

Only if we are given that its a right angle triangle with hypotenuse 10 and one side 6 (or 8), then we can conclude that the third side would be 8 (or 6). Or if we are given that all sides are integers and hypotenuse is 10, can we conclude that the other two sides (legs) would be 6/8.
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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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07 Jun 2018, 13:09
So, the take-away-property would be the Thales-Theorem...?
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In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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04 Oct 2018, 04:23
Vyshak wrote:
Perimeter of ABCD = AB + BC + CD + AD = ?

St1: The length of AB is 6 and the length of CD is 8. --> AB = 6 and CD = 8
Not Sufficient as value of BC and AD is not known.

St2: AC = diameter = 10 --> ABC and ADC are right angled triangles. Values of the sides of the quadrilateral cannot be derived.
Insufficient

Combining St1 and St2: AB = 6, CD = 8, ABC and ADC are right angled triangles
In ABC, AC = 10, AB = 6 --> BC = sqrt(10^2 - 6^2) = 8
In ADC, AC = 10, CD = 8 --> AD = sqrt(10^2 - 8^2) = 6
Perimeter = AB + BC + CD + AD = 28
Sufficient

Hello, Vyshak
thanks so much for this wholesome explanation but I don't still understand why you assumed BC=CD, is this a specific property or just because they look the same in the figure? Thanks.

Posted from my mobile device
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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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16 Oct 2018, 20:24
Bunuel wrote:

In the figure shown, quadrilateral ABCD is inscribed in a circle of radius 5. What is the perimeter of quadrilateral ABCD?

(1) The length of AB is 6 and the length of CD is 8.
(2) AC is a diameter of the circle.

Attachment:
2016-07-03_2149.png

I don't understand how we can take triangle ABC and ACD as a right angle triangle after the (1) and (2) statement. Please help..

Thanks
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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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16 Nov 2018, 04:36
Hello,

Wondering we need 2 at all?
Doesn't inscribed mean that the points touch the ends of the circle? Which means its the diameter anyway??
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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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18 Nov 2018, 03:51
1
hibobotamuss wrote:
Hello,

Wondering we need 2 at all?
Doesn't inscribed mean that the points touch the ends of the circle? Which means its the diameter anyway??

Hello

Yes, inscribed means that all the points touch the ends of the circle. But that doesnt necessarily mean that AC is diameter only.
AC and BD are the diagonals of this quadrilateral, but its quite possible that none of those two diagonals pass through the center of the circle.

Thats why second statement is needed.
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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of  [#permalink]

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18 Nov 2018, 04:06
Aahhhhhhh, makes sense. Didn't think of that :/
Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of &nbs [#permalink] 18 Nov 2018, 04:06
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