Math, asked by kukku3marhudhann7u, 1 year ago

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).

Answers

Answered by Amir93345
29



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.
Answered by shabbirmohammadshabb
0

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

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