Question

In triangle pqr, pq=13cm, pr=17cm S is mid point of QR, PT is angle bisector of angle QPR, Angle PTQ=90 degree find ST?

Answer 1

Answer:

The value of ST is 2 cm

Step-by-step explanation:

Given ΔPQR,  PQ=13cm, PR=17cm, S is mid point of QR, PT is angle bisector of angle QPR, and ∠PTQ=90° . we have to find ST.

Construction: extend line QT on the side PQ which intersect PQ at U.

In ΔPQT and ΔPRU

∠PTQ=∠PTU=90°   (given)

 PT=PT          (Common)

∠QPT=∠TPU   (Given)

By SAS rule, ΔPQT≅ΔPUT

By CPCT, PQ=PU=13 cm and QT=TU i.e T is the mid point of QR

PQ=PU+UR ⇒ 17=13+UR ⇒ UR=4 cm

In triangle UQR, T and S are the mid points of sides QU and QR.

∴ By mid-point theorem

$TS=\frac{1}{2}UR$

           = $\frac{1}{2}\times 4=2 cm$

The value of ST is 2 cm