TheCATExam
Sign in

CAT 1993 — QA Question 28

Basics of TrianglesEasy
Passage / Data

Answer the following questions based on the information given below:

ABC forms an equilateral triangle in which B is 2 km from A. A person starts walking from B in a direction parallel to AC and stops when he reaches a point D directly east of C. He, then, reverses direction and walks till he reaches a point E directly south of C.

Then D is

Answer & solution

  • A

    3 km east and 1 km north of A

  • 3 km east and 3 km north of A

  • C

    3 km east and 1 km south of A

  • D

    3 km west and 3 km north of A

Solution

Since Δ ABC is an equilateral triangle with length of the side 2 km, so its altitude will be 3 km. 

The horizontal distance of D from A = half of AB + CD = 1 + 2 = 3 kms

The vertical distance of D from A = height of the triangle ABC = √3 kms

∴ D is 3 km east and 3 km north of A.

Hence, option (b).

CAT 1993 QA Q28: Then D is — Solution | TheCATExam