Let ABCD be a quadrilateral inscribed in a circle with diameter AC, an

Question

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. \(3\)

Show Answer

Official Answer

D

andreagonzalez2k · Accepted Answer

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}

rsrighosh · Answer

Can anybody explain how to proceed ?

v12345 · Answer

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.

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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).

rsrighosh · Answer

andreagonzalez2k highly interesting approach. Thanks for sharing v12345 very well and detailed explanation. Thanks for sharing. Need to ensure I remember subtend property

thakurarun85 · Answer

How area of BCD = 1/2*BC*X

v12345 · Answer

thakurarun85 
BC extended to drop a perpendicular from D = height BCD = x
Area of triangle BCD = 1/2 * base * height = 1/2 * BC * x

100mitra · Answer

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 :-
  = 2   +  
→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

Fdambro294 · Answer

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.

bumpbot · Answer

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Let ABCD be a quadrilateral inscribed in a circle with diameter AC, an

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