ABCD is a rectangle inscribed in a circle. If the length of AB is thr

Question

ABCD is a rectangle inscribed in a circle.

144144 · Accepted Answer

i did it a bit different.

Max size for a rectangle is a square.
and minimum is when the width or length = to 0 (or almost 0)

so lets check the max and than we know that everything below the max is a possibility.

so if we know the radii is 1 - we know that if it was a square its diagonals will be 2 each.
so the area of the rectangle will be < 2*2/2 = 2

so - I, II is below 2, above 0 - than - they are ok.

ravindra88 · Answer

Rectangle  ABCD  can be broken down into 2 congruent triangles,  	riangle  ABC  &  	riangle  CDA .

area of rectangle  ABCD  = 2 * area of  	riangle  ABC

Let's draw a perpendicular from B to AC and denote the length by h.

Also,  AC = 2r = 2

Now, area of  	riangle  ABC   = (1/2)*(2)*(h) = h

area of rectangle  ABCD = 2h

Now maximum value of h can be r and minimum can be tending towards 0.

So, area of rectangle will vary between  0  and  2 .

Answer D

fluke · Answer

Sol:
Let l=length
w=wdith
d=diagonal=2
 Area=l*w=l*\sqrt{d^2-l^2}

l*\sqrt{2^2-l^2}

l*\sqrt{4-l^2}=Area

Squaring both sides;
 l^2*(4-l^2)=(Area)^2 
 4l^2-l^4=(Area)^2 
 -l^4+4l^2-(Area)^2=0

Let \hspace{2} l^2=x

-x^2+4x-(Area)^2=0

D=\sqrt{b^2-4*a*c}

D=\sqrt{4^2-4*(-1)*-(Area)^2}

D=\sqrt{16-4*(Area)^2}

In inside of the root must be greater or equal to 0 to provide valid roots of the equation.
 16-4*(Area)^2 \ge 0

4*(Area)^2 \le 16

(Area)^2 \le 4

(Area) \le 2

Only I and II are less than 2.

Ans: "D"

annmary · Answer

thank you fluke your solution is very well. +5 kudos , but i just can give you +1 kudos ;-)

other 4 kudos :+1,+1,+1,+1

ravsg · Answer

Area cannot be 3.2 because area of circle is pi r^2 i.e. 3.14, so rule out III
Even if the length is very close to the diameter i.e. 1 approx, width has to be 1/100 to make the area .01 - possible
similarly 1.9 is also possible.
=>D

144144 · Answer

thanks old friend.

shalu · Answer

area of rectangle = l*b

diagonal is 2 
also d (diagonal) = sqrt (l^2 + b^2)
l*b = l * sqrt (2 -l^2)
area is positive 
therefore (2-l^2) >0
i.e sqrt2>l
similarly sqrt2>b 
therefore area = l.b <sqrt 2.sqrt2 = 1.4 *1.4 = 1.96
therefore D is correct.

amit2k9 · Answer

Max area of the rectangle possible when diagonal = diameter = 2
Hence, l^2 + b^2 = 4. Implies l and b < 1.5 each.
Also, area of circle = 3.14 * 1^2 = 3.14. hence options C and E POE.
Now,
0<l <1.5 and 0<b<1.5. Thus area can be 0.01 and 1.9 both.
Hence D.

vipulgoel · Answer

15 seconds solution
the max area of a rectangle could be the area of squire that can be inscribed in circle with vertices on edges of circle , hence max area could be 2*2 / 2 as 2 is the max length of dignol of squire, hence D

Shubhranil88 · Answer

Trigonometric Approach

Since the radius is equal to 1, the diagonal of rectangle is equal to 2r = 2

Consider angle CAD = x

Hence CD= 2 sin x and AD= 2 cos x

Area of the rectangle ABCD= AD.CD = (2sinx)(2cosx)= 2(2sinx.cosx) = 2 sin 2x

Now sin2x has maximum value 1 & minimum value 0

Hence 0 < area (ABCD) < 2

Therefore  OPTION D

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

ABCD is a rectangle inscribed in a circle. If the length of AB is thr

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

ABCD is a rectangle inscribed in a circle. If the length of AB is thr

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards