In ΔPQR, ∠Q=90
∘
. PQ=8 cm and PR=17 cm. If the bisector of ∠P meets QR at S, then what is the length (in cm) of SR?
- A12.4
- B9.6
- C10.2
- D8.4
Solution & Step-by-step Explanation
In right-angled ΔPQR with ∠Q=90
∘
:
By Pythagoras theorem:
QR=
PR
2
−PQ
2
=
17
2
−8
2
=
289−64
=
225
=15 cm
Since PS is the angle bisector of ∠P, by the Angle Bisector Theorem, the ratio of the segments of the opposite side equals the ratio of the other two sides:
SR
QS
=
PR
PQ
=
17
8
Let QS=8k and SR=17k.
Given that QR=QS+SR=15 cm:
8k+17k=15
25k=15
k=
25
15
=
5
3
=0.6
Therefore, the length of SR is:
SR=17k=17×0.6=10.2 cm
∘
:
By Pythagoras theorem:
QR=
PR
2
−PQ
2
=
17
2
−8
2
=
289−64
=
225
=15 cm
Since PS is the angle bisector of ∠P, by the Angle Bisector Theorem, the ratio of the segments of the opposite side equals the ratio of the other two sides:
SR
QS
=
PR
PQ
=
17
8
Let QS=8k and SR=17k.
Given that QR=QS+SR=15 cm:
8k+17k=15
25k=15
k=
25
15
=
5
3
=0.6
Therefore, the length of SR is:
SR=17k=17×0.6=10.2 cm