The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is

(A) (a + b + c) 2     

(B) 0  

(C) a + b + c  

(D) abc

Slide45.JPG

Slide46.JPG


Transcript

Question 21 The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is (A) (a + b + c)2 (B) 0 (C) a + b + c (D) abc Given vertices (a, b + c), (b, c + a) and (c, a + b) Here x1 = a , y1 = b + c x2 = b , y2 = c + a x3 = c , y3 = a + b Now, Area of triangle = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] = 1/2 [ a([c + a] – [a + b]) + b([a + b] – [b + c]) + c([b + c] – [c + a]) ] = 1/2 [ a(c − b) + b(a − c) + c(b − a) ] = 1/2 [ ac − ab + ab − bc + bc − ac] = 1/2 × 0 = 0 square units So, the correct answer is (B)

Go Ad-free
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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.