NCERT Exemplar - MCQ

Chapter 7 Class 10 Coordinate Geometry
Serial order wise

## (A) 5   (B) 12  (C) 11   (D) 7 + √5

### Transcript

Question 9 The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is (A) 5 (B) 12 (C) 11 (D) 7 + √5 Let A (0, 4), B (0, 0) and C (3, 0) Thus, Perimeter of triangle = AB + BC + AC Finding AB AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 0−0)2+(0 −4)2) = √(02+(−4)2) = √16 = √(4^2 ) = 4 units Finding AC Here, A (0, 4) and C (3, 0) AC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √((3−0)2+(0 −4)2) = √(32+(−4)2) = √(9+ 16) = √25 = √(5^2 ) = 5 units Finding BC Here, B (0, 0) and C (3, 0) BC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √((3−0)2+(0 −0)2) = √(32+(0)2) = √(9+0) = √9 = √(3^2 ) = 3 units Thus, AB = 4, AC = 5, BC = 3 Therefore, Perimeter of triangle = AB + BC + AC = 4 + 3 + 5 = 12 units So, the correct answer is (B)

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.