The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

(A) 5   (B) 12  (C) 11   (D) 7 + √5

 

Slide13.JPG

Slide14.JPG
Slide15.JPG Slide16.JPG

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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)

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.