Question

In the given figure, PT is parallel to QR and QT is parallel to RS. Show that ar( triangle PQR) is equal to ar( triangle QTS).

Answer 1




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.

Answer 2

Step-by-step explanation:

PQRS is a square and SRT is an equilateral triangle. To prove: (i) PT =QT and (ii) ∠TQR = 15° Now, PQRS is a square: PQ = QR = RS = SP …… (i) And ∠ SPQ = ∠ PQR = ∠ QRS = ∠ RSP = 90° Also, △ SRT is an equilateral triangle: SR = RT = TS …….(ii) And ∠ TSR = ∠ SRT = ∠ RTS = 60° From (i) and (ii) PQ = QR = SP = SR = RT = TS ……(iii) From figure, ∠TSP = ∠TSR + ∠ RSP = 60° + 90° = 150° and ∠TRQ = ∠TRS + ∠ SRQ = 60° + 90° = 150° ∠ TSR = ∠ TRQ = 150° ………………… (iv) By SAS congruence criterion, Δ TSP ≃ Δ TRQ We know, corresponding parts of congruent triangles are equal So, PT = QT Proved part (i). Now, consider ΔTQR. QR = TR [From (iii)] Δ TQR is an isosceles triangle. ∠QTR = ∠TQR [angles opposite to equal sides] Sum of angles in a triangle = 180° ∠QTR + ∠ TQR + ∠TRQ = 180° 2∠TQR + 150° = 180° [From (iv)] 2 ∠TQR = 30° ∠TQR = 15° Hence proved part (ii).Read more on Sarthaks.com - https://www.sarthaks.com/609505/in-figure-pqrs-is-a-square-and-srt-is-an-equilateral-triangle-prove-that-i-pt-qt-ii-tqr-15