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# 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 R and Q (B-R-Q-D). then find the area of triangle PQR. i didnt understand ur xplanation plzz xplain again....

147 Points
11 years ago

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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