Question

Actually whatever I'm typing is the solution of a question that one random guy asked......so the question was .....ABCD is a square with AC as its diagonal and M as mid point of AB and a line from point D meets AB at point M and AC and DM intersects each other at point O , so find the ratio of ODC to area of square ....so here here I go with the solution forcthe same .... Now , Let the side of the given square be x . so area=x² now as AC has divided the square ABCD into 2 triangles of equal ares and each triangle is half of area of ABCD so area of triangle ADC =x²/2 , now areaADC= areaAOD +areaODC so firstly lets find the area of triangle AOD , for that concern firstly find out the value of DM . so , use pythagoras theorem over here, so x²+(x/2)²= x²+x²/4 =DM2...so DM =( 5^1/2)x /2 now take the triangle ADM , its area equals to( AD×AM)/2 means x²/4 abd area of ADM could also be (OA ×DM )/2 so( OA×(5^1/2)x/2)/2 =x²/4 so OA = (5^1/2)x/5 (after rationalisation) so now come back to triangle AOD , so do construction over here....now draw a median DE perpendicular to OA so Area of AOD =( DE×AO)/2 so here use pythagoras theorem again to find DE , therefore, AE= OA/2 =(5^1/2)x/10 , therefore, DE² = AD²-AE² so now , DE=( ( 19^1/2))x/20^1/2 , now area of triangle ODC= areaADC-areaAOD , so area ODC =x²/2 -(19^1/2)x/20 so area of ODC = (x²(10-19^1/2))/20 now ratio =((x²(10-19^1/2))/20) /x²........so final answer = (10-19^1/2)/20 or you can say (10-19^1/2):20 .....it is a very important maths question....and whoever is watching this question with solution is licky and plz write it in fair notebook​

Answer 1

Answer:

ok Busch r6std8dg8d8t

Answer 2

Dude what are u tryin?.?.?.??..??

I am surprised at the fact that you could write so much......

Just mark me ask the brainiest cause I can’t think of more words than this....

Plzzzz bro