PQRS is quadrilater PR and QS intersect at T . Prove that ar(PTQ)*ar(STR)=ar(PST)*ar(QTR).

Question

PR&QS are diagonals.

Answer 1

???????????? yu meen a 2-d shape with 4 sides?????????

????? then sumthun tu du with wot get if draw diagonals

with 4 sides, hav 2 possabel diagonals

Giv us a klue bout wot yu look for

Answer 2

Question: PQRS is quadrilater PR and QS intersect at T . Prove that ar(PTQ)*ar(STR)=ar(PST)*ar(QTR).

I assume that ar(PTQ) means the area of the triangle PTQ.

The diagonals QS and PR intersect at T.

There are 4 angles at this point of intersection. They are both opposite and complementary.

Let the four angles be T, T', T and T', where T = 180 - T'.

From this we have sin(T) = sin(T').

The area of a triangle ABC is given by Area = (1/2)a.b.sin(C)

ar(PTQ) = (1/2)PT.QT.sin(T)

ar(STR) = (1/2)ST.RT.sin(T)

ar(PST) = (1/2)PT.ST.sin(T')

ar(QTR) = (1/2)QT.RT.sin(T')

ar(PTQ)*ar(STR) = (1/4)PT.QT.ST.RT.sin^2(T) = (1/4)PT.QT.RT.ST.sin^2(T)

ar(PST)*ar(QTR) = (1/4)PT.ST.QT.RT.sin^2(T') = (1/4)PT.QT.RT.ST.sin^2(T')

And sin(t) = sin(T'),

Hence ar(PTQ)*ar(STR) = ar(PST)*ar(QTR)