PLEASE DISCUSS THIS QUESTION: In a traingle ABC, medians AD and BE are drawn; If AD=4 , angle DAB=30 degree, and angle ABE= 60 degree , then what is the area of traingle ABC? options: a)8/3 b)16/3 c)32/3 d)64/3
AD, BE are medians let they intersect at point O.
now any median divides the area of a triangle into 2 equal parts so AREA ABD= AREA ADC
now taking triangle AOD, angle DAB therefore angle OAB= 30, Angle ABE therefore angle ABO= 60
thus angle AOB= 90, Triangle is a right angled triangle…
thus AREA ABD = 1/2 * AD * BO, here AD=4, BO= AO * tan 30 degree (from triangle AOB)
area ABC = 2 * 1/2 * AD * AO * tan 30 = AD * AO * 1/ SQUARE ROOT 3
now AO = 2/3 * AD ( as when two medians intersect they are divided in the ratio of 2: 1)
thus multiply all this to get ur answer
NOTE: i think there needs to be square root 3 (according to my solution it will come)
Post Comments
vasudevakh said – Mon, 19 Oct 2009 16:34:07 -0000 ( Flag Edit Link )
perfectly right, Answer is 32/3 square root of 3
deepa mittal said – Fri, 30 Oct 2009 05:05:57 -0000 ( Flag Edit Link )
That’s correct.
AD = 4, AO = 8/3
Area of triangle = AD * AO * 1/ SQUARE ROOT 3 = 8/3 * 4 * 1/(square root of 3)
= 32/3*(square root of 3)