Discussion

Explanation:

Since S is the midpoint of QR, A(∆PSR)

= A(∆PQR)/2

= 5.8/2 = 2.9 sq.cm

Now A(∆PSR) = A(∆PSX) + A(∆SXR)

Also, A(∆RTX) = A(∆TSX) + A(∆SXR)

Now, triangles PSX and TSX lie within the same parallel lines – SX and PT – and hence, have the same area.

∴ A(∆PSX) = A(∆TSX)

∴ A(∆PSR) = A(∆RTX) = 2.9 sq.cm

Hence, option (a).

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