I am unable to view the attachment. Can you copy into word document and attach?sairamGmat wrote:
If A is the center of the circle shown above and AB=BC=CD, what is the value of x?
(A) 15
(B) 30
(C) 45
(D) 60
(E) 75
OA is B
Any explanation on how the OA answer came? Thank you
kvcpk wrote:I am unable to view the attachment. Can you copy into word document and attach?sairamGmat wrote:
Attached...Thanks.....
If A is the center of the circle shown above and AB=BC=CD, what is the value of x?
(A) 15
(B) 30
(C) 45
(D) 60
(E) 75
OA is B
Any explanation on how the OA answer came? Thank you
can we look at it this way, since the diagonals are perpendicular bisectors,angle AOB= 90.kvcpk wrote:Complete the picture ABCD by drawing the lines BC and CD.
Now in quadrilateral ABCD, AB=BC=CD (Given)
AB=BC=CD=DA (because AB=DA -> radius)
Let the intersection point of the diagonals be O
Hence, quadrilateral ABCD is a rhombus. We know the the diagonals of a rhombus are perpendicular bisectors.
Hence angle(AOB) = BOC=COD=DOA=90
now, we know that angle(ABO) is x.
Traingle ABD is isoceles. hence angle(ADB)=x
Let angle OBC=y.
Traingle BCD is isoceles. hence angle(BDC)=y
here is an important theorem. Angle subtended by an arc of a circle at the centre is twice the angle it subtends on the same side of the circle.
CD arc is suspending y degrees at B. So angle(DAC) = 2y
If you look at arc BC, then For the same reason we get that angle(BAC)=2y
Now in triangle ABD, 4y+2x=180
y+2x=90 .. stmt1
In triangle AOB, x+2y+90=180
x+2y=90 ... stmt2
By solving both we get x=y=30
Hope this helps!! let me know if you have issues understanding this.
It sure does... I was just about to post that method, because I thought using circle theorem would be toomuch on GMAT.raunakrajan wrote: can we look at it this way, since the diagonals are perpendicular bisectors,angle AOB= 90.
AB=BC=AC therefore TRIANGLE ABC is equilateral.
if AOB=90, BAO=60 AB0 has to be 30deg
does it make sense?
yea it would be too much to apply fr such a problem and probably waste time.kvcpk wrote:It sure does... I was just about to post that method, because I thought using circle theorem would be toomuch on GMAT.raunakrajan wrote: can we look at it this way, since the diagonals are perpendicular bisectors,angle AOB= 90.
AB=BC=AC therefore TRIANGLE ABC is equilateral.
if AOB=90, BAO=60 AB0 has to be 30deg
does it make sense?
AB=BC=AC therefore TRIANGLE ABC is equilateral.
The diagonals of a rhombus are angular bisectors. So x will be 60/2 = 30
:
shashank.ism wrote:See its a very simple problem...solve it this way ...
since AB = BC also AB = AC(radius of circle).. so AB = BC = AC hence ABC is equilateral
so <BAC = 60`
Let intersection of BD and AC = P
so <BPC = 90`Hence x` = <BPA = 180-90-60 = 30` Ans B.[/quote
A silly question to ask...should we assume they intersect at 90 deg?
There is no right angle stated there right? Shouldnt we prove its 90 deg first before we concluded is 30-60-90.
Since A is the center of the circle AB=AD. Also, it's given that AB=BC=CD. Therefore, AB=BC=CD=AD.sairamGmat wrote:
If A is the center of the circle shown above and AB=BC=CD, what is the value of x?
(A) 15
(B) 30
(C) 45
(D) 60
(E) 75
OA is B
Any explanation on how the OA answer came? Thank you
