




Subscribe to our Youtube Channel - https://you.tube/teachoo
Last updated at May 29, 2018 by Teachoo
Transcript
Ex 4.3,1 Find area of the triangle with vertices at the point given in each of the following: • (1, 0), (6, 0), (4, 3) The area of triangle is given by ∆ = 12 x1y11x2y21x3y31 Here, x1 = 1 , y1 = 0 x2 = 6 ,y2 = 0 x3 = 4 ,y3 = 3 ∆ = 12 101601431 = 12 1 0131−0 6141+1 6043 = 12 (1(0 – 3) – 0(6 – 4) + 1 (18 – 0)) = 12 (1( – 3) + 0 + 1 (18) ) = 12 [ – 3 + 18 ] = 152 Thus, the required area of triangle is 𝟏𝟓𝟐 square units Ex 4.3,1 Find area of the triangle with vertices at the point given in each of the following: (ii) (2, 7), (1, 1), (10, 8) The area of triangle is given by ∆ = 12 x1y11x2y21x3y31 Here, x1 = 2 , y1 = 7 x2 = 1 , y2 = 1 x3 = 10 , y3 = 8 ∆ = 12 2711111081 = 12 2 1181−7 11101+1 11108 = 12 ( 2(1 – 8) – 7(1 – 10) + 1 (8 – 10) ) = 12 (2( – 7) – 7( – 9) + 1 (– 2) ) = 12 ( – 14 + 63 – 2 ) = 12 [– 16 + 63] = 472 Thus, the required area of triangle is 𝟒𝟕𝟐 square units Ex 4.3,1 Find area of the triangle with vertices at the point given in each of the following: (iii) (−2, −3), (3, 2), (−1, −8) The area of triangle is given by ∆ = 12 x1y11x2y21x3y31 Here, x1 = – 2, y1 = – 3 x2 = 3 , y2 = 2 x3 = – 1, y3 = – 8 ∆ = 12 −2−31321−1−81 = 12 −2 21−81−(−3) 31−11+1 32−1−8 = 12 [ – 2(2 – ( – 8)) + 3(3 – ( – 1)) + 1 ( – 24 – ( –2) ] = 12 [ – 2(2 + 8) + 3(3 + 1) + 1 (– 24 + 2) ] = 12 [ – 2(10) + 3 (4) + 1 ( – 22) ] = 12 ( – 20 – 22 + 12) = 12 (− 30) = – 15 Since the area of triangle is always positive, the required area of triangle is 15 square units
About the Author