Question

In figure, PS, SQ, PT and TR are 4cm, 1cm, 6cm and 1.5cm respectively. Prove that ST||QR. Also find area of ΔPST/area of trapezium QRTS.​

Answer 1

Step-by-step explanation:

1. ratio of PS/SQ=PT/TR
2. Then by BPT therom ST||QR

Answer 2

The area of Δ PST by the area of trapezium QRTS = 16/9

Step-by-step explanation:

See the attached figure.

In the given figure, PS = 4 cm,  SQ = 1 cm, PT = 6 cm and TR = 1.5 cm.

Now, we have $\frac{PS}{SQ} = \frac{PT}{TR} = \frac{4}{1} = \frac{6}{1.5} = 4$ ............. (1)

Hence, in Δ PQR, line ST divides the two sides of Δ PQR, PQ and PR in the same ratio {as given by the equation (1)}and, therefore, ST ║ QR.

Now, Δ PST and Δ PQR are similar triangles. {Since ST ║ QR and ∠ PST = ∠ PQR and ∠ PTS = ∠ PRQ}

So, ST : QR = PS : PQ = PT : PR = 4 : 5

Let, ST = 4x and QR = 5x.

Now, draw a perpendicular on QR from the vertex P and that is PV, then PU will also be perpendicular to ST.

Now, since ST ║ QR, so, PS : SQ = 4 : 1 = PU : UV

Let, PU = 4y and UV = y,

So, the area of Δ PST = $\frac{1}{2} \times (PU) \times (ST) = \frac{1}{2}(4x)(4y) = 8xy$

And, the area of trapezium QRTS = $\frac{1}{2} \times (ST + QR) \times (UV) = \frac{1}{2}(4x + 5x)(y) = 4.5xy$

Therefore, the area of Δ PST by the area of trapezium QRTS = $\frac{8xy}{4.5xy} = \frac{16}{9}$ (Answer)