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Mathematics 29 Online
OpenStudy (shubhamsrg):

geometry ques.. image coming up..

OpenStudy (shubhamsrg):

OpenStudy (shubhamsrg):

@eliassaab @mukushla @ganeshie8

OpenStudy (mathteacher1729):

Is this from the Entrance Examination for CMI BSc (Mathematics & Computer Science) May 2011from the Chennai Mathematical Institute? http://www.cmi.ac.in/admissions/sample-qp/ugmath2011.pdf ?

OpenStudy (shubhamsrg):

yes..am aiming for it next year..

OpenStudy (mathteacher1729):

There are some interesting questions on there. I wish you the best of luck, it's good that you're preparing now! :)

OpenStudy (shubhamsrg):

yes,,altough most questions werent that difficult..but true,nice bunch of problems..

OpenStudy (anonymous):

Since the opposite angles are right angles the quadrilateral can either be a square or a rectangle.The proof is in the picture.

OpenStudy (anonymous):

And AAS congruency rule is Angle Angle side that is if the corresponding two angles and side of two triangles are equal they are said to be congruent.

OpenStudy (shubhamsrg):

buddy that assumption is wrong,, 2 angles 90 doesnt mean other 2 will also be 90

OpenStudy (shubhamsrg):

this can be concluded though ,since opposite angles 90,,i.e. sum to 180,,its a cyclic quad,,not so sure if that'd help..

OpenStudy (anonymous):

If the angles are not opposite the assumption is wrong but they are opposite bro.They can only be 90 degrees.|dw:1344697497559:dw|

OpenStudy (mathteacher1729):

These triangles share a hypotenuse. You can stitch together two 3,4,5 triangles such that they share the hypotenuse, but the 3 and 4 are adjacent, not opposite.

OpenStudy (shubhamsrg):

well how about this : |dw:1344698175488:dw| which is exactly what @mathteacher1729 says..

OpenStudy (shubhamsrg):

anyways,,i have got the solution..thanks all for the efforts by the way..hold on,,i'll post the solution..

OpenStudy (anonymous):

I called my maths teacher and verified it.He says its right.

OpenStudy (shubhamsrg):

let AM=m , CN=n , DM=a , MN=b , NB =c <ADM> = y and <ABM>=x now <CDN>=90-y => <DCN> = y also <CBM>=90-x => <NCB>=x tan x = m/(b+c) = c/n tany = m/a = (a+b)/n from these 2 eqns ,we have mn = c(b+c) = a(a+b) => (b+c)/(a+b) = a/c ......eqn (1) subtracting 1 from both sides in eqn(1) ,, =>(c-a)/(a+b) =(a-c)/(a+c) => (c-a)(a+c) = (a-c)(a+b) =>(c-a)(a+c) - (a-c)(a+b) =0 =>(c-a)(a+c+a+b) =0 well 2a+b+c is ofcorse not equal to 0 only thing which can be 0 here is c-a this c-a=0 or c=a.. hence proved!! phew!

OpenStudy (shubhamsrg):

lol i missed the fig,,1 min.,,

OpenStudy (shubhamsrg):

|dw:1344698950393:dw|

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