Find the area of triangle abc with vertices a(0,-1), b(2,1) and c(0,3). also, find the area of triangle formed by joining midpoints of its sides. show that the ratio of area of two triangle id 4:1 .
Answers
Distance between the points a and b = √(2-0)²+(1-(-1))²
= √2²+2²
= √8 = 2√2
Distance between the points b and c = √(2-0)²+(1-3)²
= √2²+2²
= √8 = 2√2
Distance between the points a and c = | 3-(-1) |
= 4
Area of triangle = 1/2 | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) |
= 1/2 | 0(1-3)+2(3+1)+0(-1-1)
= 1/2 |8|
= 4
Let P,Q and R mid points of side AB, BC and AC
Coordinates of P = (1,0) {using the formulae of midpoint }
Coordinates of Q = (1,2)
Coordinates of R = (0,1)
The area of triangle PQR = 1/2 | 1(2-1)+1(1-0)+0(0-2) |
= 1/2 | 2 |
= 1
Ratio of area of triangle ABC to the area of triangle PQR = 4:1
Hope it helps u...........................
Answer:
hope it helps u........
