ln the figure ,PT||QR and QT||RS show that ar (∆PQR )=ar(∆QTS)
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Hi friend ...
Given that: PT || QR and QT || RS
To Prove: ar(ΔPQR)=ar(ΔQTS)
Proof:
Since triangle PQR and TQR are on the same base QR and between the parallels PT and QR. Therefore,
ar(ΔPQR)=ar(ΔTQR) ..... (1)
Similarly, triangle QTS and TQR are on the same base QT and between the parallels QT and RS. So,
ar(ΔQTS)=ar(ΔTQR) ..... (2)
From (1) and (2), we have,
ar(ΔPQR)=ar(ΔQTS)
Hence Proved.
hope this helps u.....
:)
Given that: PT || QR and QT || RS
To Prove: ar(ΔPQR)=ar(ΔQTS)
Proof:
Since triangle PQR and TQR are on the same base QR and between the parallels PT and QR. Therefore,
ar(ΔPQR)=ar(ΔTQR) ..... (1)
Similarly, triangle QTS and TQR are on the same base QT and between the parallels QT and RS. So,
ar(ΔQTS)=ar(ΔTQR) ..... (2)
From (1) and (2), we have,
ar(ΔPQR)=ar(ΔQTS)
Hence Proved.
hope this helps u.....
:)
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