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In the figure shown, quadrilateral ABCD is inscribed in a circle of
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03 Jul 2016, 10:51
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73% (01:13) correct 27% (01:21) wrong based on 483 sessions
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In the figure shown, quadrilateral ABCD is inscribed in a circle of
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03 Jul 2016, 11:11
Given: Radius = 5 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
Answer: C




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In the figure shown, quadrilateral ABCD is inscribed in a circle of
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Updated on: 04 Jul 2016, 20:25
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, 23:29.
Last edited by hsbinfy on 04 Jul 2016, 20:25, edited 1 time in total.



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In the figure shown, quadrilateral ABCD is inscribed in a circle of
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03 Jul 2016, 23:40
Vyshak wrote: Given: Radius = 5 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
Answer: C 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
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04 Jul 2016, 00:33
hsbinfy wrote: Vyshak wrote: Given: Radius = 5 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
Answer: C 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
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10 Jul 2016, 07:36
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
Answer: C



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In the figure shown, quadrilateral ABCD is inscribed in a circle of
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10 Jul 2016, 19:04
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=\(\sqrt{AC^2+AB^2}\)=\(\sqrt{10^26^2}\)=8 Accordingly we can find AD=6 So,perimeter of quadrilateral ABCD=AB+BC+CD+AD=6+8+8+6=28, SufficientCorrect Answer C
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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of
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27 Apr 2017, 15:40
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
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29 Sep 2017, 05:47
Hi,
A quadrilateral inside a cirlce is called a cyclic quadrilateral.
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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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of
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29 Sep 2017, 06:14



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Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of
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06 Dec 2017, 05:21
(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 6810?
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
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06 Dec 2017, 23:38
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 6810?
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 6810 triangle. It could also be 5√25√210 triangle OR it could be 2√54√510 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
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07 Jun 2018, 14:09
So, the takeawayproperty would be the ThalesTheorem...?




Re: In the figure shown, quadrilateral ABCD is inscribed in a circle of &nbs
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07 Jun 2018, 14:09






