in figure s and t are the points on the side PQ and PR respectively of triangle PQR such that PT = 4cm TR =4 cm and ST is parallel to QR. find the ratio of the areas of triangle PST and PQR
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Given:
ST || QR
PT= 4 cm
TR = 4cm
In ΔPST and ΔPQR,
∠SPT = ∠QPR (Common)
∠PST = ∠PQR (Corresponding angles)
ΔPST ∼ ΔPQR (By AA similarity criterion)
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
ar(∆PST) /ar(∆PQR)= (PT)²/(PR)²
ar(∆PST) /ar(∆PQR)= 4²/(PT+TR)²
ar(∆PST) /ar(∆PQR)= 16/(4+4)²= 16/8²=16/64= 1/4
Thus, the ratio of the areas of ΔPST and ΔPQR is 1:4.
HOPE THIS WILL HELP YOU...
ST || QR
PT= 4 cm
TR = 4cm
In ΔPST and ΔPQR,
∠SPT = ∠QPR (Common)
∠PST = ∠PQR (Corresponding angles)
ΔPST ∼ ΔPQR (By AA similarity criterion)
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
ar(∆PST) /ar(∆PQR)= (PT)²/(PR)²
ar(∆PST) /ar(∆PQR)= 4²/(PT+TR)²
ar(∆PST) /ar(∆PQR)= 16/(4+4)²= 16/8²=16/64= 1/4
Thus, the ratio of the areas of ΔPST and ΔPQR is 1:4.
HOPE THIS WILL HELP YOU...
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