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805+ (Hard)|   Geometry|      
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v12345
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Let ABCD be a quadrilateral inscribed in a circle with diameter AC, an 17 Sep 2021, 03:32
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

34% (03:21) correct 66% (02:55) wrong based on 41 sessions

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Let ABCD be a quadrilateral inscribed in a circle with diameter AC, and let E be the foot of perpendicular from D onto AB, as shown in the figure below. If AD = DC and the area of quadrilateral ABCD is 24 \(cm^2\), find the length of DE.
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A. \(3\sqrt{6}\)
B. \(2\sqrt{2}\)
C. \(4\sqrt{2}\)
D. \(2\sqrt{6}\)
E. \(3\)
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D
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andreagonzalez2k
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Let ABCD be a quadrilateral inscribed in a circle with diameter AC, an 17 Sep 2021, 17:37
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rsrighosh
Can anybody explain how to proceed ?
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Quick approach:
as we don't really know angle BAC we can suppose that the answer does not depend on where the point B is, so we can move it to the left to be in the same line as D and O, so the quadrilateral ABCD is turned into a square being DE one of its sides.

\(A = DE^2 = 24\) -> \(DE = \sqrt{24} = 2\sqrt{6}\)
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rsrighosh
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Let ABCD be a quadrilateral inscribed in a circle with diameter AC, an 17 Sep 2021, 08:13
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Can anybody explain how to proceed ?
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v12345
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Let ABCD be a quadrilateral inscribed in a circle with diameter AC, an 17 Sep 2021, 11:34
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Let r be the radius of the circle with center O.
∠ADC = ∠ABC = 90° ( Diameter AC subtends right angle at circumference)
AD = DC, ∠ACD = 45°

AD subtends equal angles at circumference,
∠ACD = 45°, so ∠ABD = 45°
Then DE = EB = x.
Attachment:
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CaptureAns.PNG [ 44 KiB | Viewed 6268 times ]

area of quadrilateral ABCD = area of AED + area of EBD + area of BCD = 24
\(\frac{1}{2}(AB - x)*x + \frac{1}{2}x^2 + \frac{1}{2}BC*x = 24\)
\(\frac{1}{2}(AB+BC)*x= 24\)
\(x= \frac{48}{(AB + BC)}\)

Also,
area of quadrilateral ABCD = area of ABC + area of ACD = 24
\(\frac{1}{2}AB*BC + \frac{1}{2}AC*OD= 24\)
\(\frac{1}{2}AB*BC + r^2= 24\)
\(AB*BC = 2(24 - r^2)\)

In triangle ABC,
\(AB^2 + BC^2 = AC^2 = 4r^2\)
\((AB + BC)^2 - 2*AB*BC= 4r^2\)
\((AB + BC)^2 = 4r^2 + 4(24 - r^2) = 96\)
\(AB + BC = 4\sqrt{6}\)

\(x= \frac{48}{(AB + BC)} = \frac{48}{4\sqrt{6}} = 2\sqrt{6}\)
Hence, OA is (D).
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Let ABCD be a quadrilateral inscribed in a circle with diameter AC, an 17 Sep 2021, 23:19
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andreagonzalez2k highly interesting approach. Thanks for sharing
v12345 very well and detailed explanation. Thanks for sharing. Need to ensure I remember subtend property :)
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Let ABCD be a quadrilateral inscribed in a circle with diameter AC, an 18 Sep 2021, 04:38
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v12345
Let r be the radius of the circle with center O.
∠ADC = ∠ABC = 90° ( Diameter AC subtends right angle at circumference)
AD = DC, ∠ACD = 45°

AD subtends equal angles at circumference,
∠ACD = 45°, so ∠ABD = 45°
Then DE = EB = x.
Attachment:
CaptureAns.PNG

area of quadrilateral ABCD = area of AED + area of EBD + area of BCD = 24
\(\frac{1}{2}(AB - x)*x + \frac{1}{2}x^2 + \frac{1}{2}BC*x = 24\)
\(\frac{1}{2}(AB+BC)*x= 24\)
\(x= \frac{48}{(AB + BC)}\)

Also,
area of quadrilateral ABCD = area of ABC + area of ACD = 24
\(\frac{1}{2}AB*BC + \frac{1}{2}AC*OD= 24\)
\(\frac{1}{2}AB*BC + r^2= 24\)
\(AB*BC = 2(24 - r^2)\)

In triangle ABC,
\(AB^2 + BC^2 = AC^2 = 4r^2\)
\((AB + BC)^2 - 2*AB*BC= 4r^2\)
\((AB + BC)^2 = 4r^2 + 4(24 - r^2) = 96\)
\(AB + BC = 4\sqrt{6}\)

\(x= \frac{48}{(AB + BC)} = \frac{48}{4\sqrt{6}} = 2\sqrt{6}\)
Hence, OA is (D).
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How area of BCD = 1/2*BC*X
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Let ABCD be a quadrilateral inscribed in a circle with diameter AC, an 18 Sep 2021, 20:27
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thakurarun85

How area of BCD = 1/2*BC*X
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thakurarun85
BC extended to drop a perpendicular from D = height BCD = x
Area of triangle BCD = 1/2 * base * height = 1/2 * BC * x
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Let ABCD be a quadrilateral inscribed in a circle with diameter AC, an 18 Sep 2021, 22:32
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Correct Option : D


First, we have that BC || ED , which is obvious from the information.
Next is that since AD=DC , BD is the angle bisector of ∠BAC which gives BED an isosceles triangle and DE=BE.
Now draw a perpendicular from C to DE and let it intersect at F .
We have that ΔAED≅ΔDFC from RHS , and this gives all the necessary piece of info in the picture, also noting that CFEB is a rectangle.
This total areas of the 2 triangles and the rectangle gives the area of the quadrilateral

From here :-
[ABCD] = 2 [ΔAED] + [CFEB]
→24 = x(x+y) + y(x+y)
→24 = (x+y)^2
→(x+y) = DE = √24

Hence DE^2 = 24 .

PS : This is not my solution, found while serching for solution on web, credit goes to the person, who has solved it, sharing as a learning for better of all.

Posted from my mobile device
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Let ABCD be a quadrilateral inscribed in a circle with diameter AC, an 16 Oct 2021, 19:26
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Thank you for this.

That first piece of information is key.

If from the same Line BA

A line segment ED and a line segment BC both “depart” or extend from that Line BA from the same Angle (90 degrees) ———-> then ED and BC will be parallel.

Hard question.


100mitra
Correct Option : D


First, we have that BC || ED , which is obvious from the information.
Next is that since AD=DC , BD is the angle bisector of ∠BAC which gives BED an isosceles triangle and DE=BE.
Now draw a perpendicular from C to DE and let it intersect at F .
We have that ΔAED≅ΔDFC from RHS , and this gives all the necessary piece of info in the picture, also noting that CFEB is a rectangle.
This total areas of the 2 triangles and the rectangle gives the area of the quadrilateral

From here :-
[ABCD] = 2 [ΔAED] + [CFEB]
→24 = x(x+y) + y(x+y)
→24 = (x+y)^2
→(x+y) = DE = √24

Hence DE^2 = 24 .

PS : This is not my solution, found while serching for solution on web, credit goes to the person, who has solved it, sharing as a learning for better of all.

Posted from my mobile device
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Let ABCD be a quadrilateral inscribed in a circle with diameter AC, an 18 Sep 2024, 09:48
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