Question : In ΔPQR, PS is the internal bisector of $\angle P$ meeting QR at S, PQ = 16 cm, PR = 22.4 cm, QR = 9.6 cm. The length of SR (in cm) is:
Option 1: 4
Option 2: 4.4
Option 3: 6
Option 4: 5.6
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
Correct Answer: 5.6
Solution :
Given,
PS is the internal angle bisector of $\angle P$
We know that The angle bisector of a triangle divides the opposite sides into two segments that are proportional to the other side of the triangle
⇒ $\frac{PQ}{PR} = \frac{QS}{SR}$
⇒ $\frac{16}{22.4} = \frac{QS}{SR}$
⇒ $\frac{QS}{SR} = \frac{5}{7}$
It is given that QR = 9.6 cm
⇒ QS + SR = 9.6 cm
⇒ 5 unit + 7 unit = 9.6 cm
⇒ 12 unit = 9.6 cm
⇒ 1 unit = 0.8 cm
⇒ QS = (5 × 0.8) cm and SR = (7 × 0.8) cm
⇒ QS = 4 cm and SR = 5.6 cm
Hence, the correct answer is 5.6 cm.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
