in a rectangle ABCD ,P is midpoint of seg.AB.S & T are points of trisection of seg. DC . if area of the rectangle is 70 sq.units. & a triangle is drawn ie.triangle PST and a diagonal bd of rectangle cuts this triangle in points Rand Q (B-R-Q-D). then find the area of triangle PQR.i didnt understand ur xplanation plzz xplain again....

Dear ajinkya

to find the area of triangle PRQ ,we should know the co ordinates of point P ,Q ,R

so our task is to find the co-ordinate of P,Q,R

let side of rectangle is a and b

 then area of rectangle ABCD =a*b  =70(given) .................1

so P will be (a/2,0)    (given it is a mid point of AB)

Point S will be (a/3,b) and T will be (2a/3,b)

Point Q is the intersection of line  BD and SP

 equation of line BD :  x/a + y/b =1

equation of line SP  : y =-6b/a (x-a/2)  { can be easly derived by equation of line from given 2 points)

 solve these line  and find Q   ,Q(2a/5 ,3b/5)

Point R is the intersection of line  BD and TP

 equation of line BD :  x/a + y/b =1

equation of line TP  : y =6b/a (x-a/2) { can be easly derived by equation of line from given 2 points)

 solve these line  and find R   ,Q(4a/7 ,3b/7)

now we know all three points P ,Q and R

use the formula of triangle and find area ,are will come in terms of ab ,put ab =70 from equation 1 , area will come 3 sq unit

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