Question : PQRA is a rectangle, AP = 22 cm and PQ = 8 cm. $\triangle$ABC is a triangle whose vertices lie on the sides of PQRA such that BQ = 2 cm and QC = 16 cm. Then the length of the line joining the midpoints of the sides AB and BC is:
Option 1: $4\sqrt2$ cm
Option 2: $5$ cm
Option 3: $6$ cm
Option 4: $10$ cm
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
Correct Answer: $5$ cm
Solution :
Given: PQRA is a rectangle, AP = 22 cm and PQ = 8 cm. $\triangle$ABC is a triangle whose vertices lie on the sides of PQRA such that BQ = 2 cm and QC = 16 cm.
Let D and E be the two midpoints of BC and AB, respectively.
PQRA is a rectangle, PA = QR = 22 cm, PQ = AR = 8 cm and QC = 16 cm.
So, CR = (22 – 16) = 6 cm
From $\triangle$ACR,
AC = $\sqrt{8^2+6^2}$
⇒ AC = 10 cm
Since D and E are the two midpoints of BC and AB, respectively,
So, DE = $\frac{AC}{2}$
⇒ DE = $\frac{10}{2}$
⇒ DE = 5 cm
Hence, the correct answer is 5 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 “”.
