Maths ar. of triangles shubham n group

LEARNING OBJECTIVES
 You will be able to :-
1. Calculate the area of a triangle when the co-
ordinates of vertices are known .
2. Calculate the area of a triangle in ease .

• Area of triangle =
½ [x1(y2-y3) +x2(y3-y1) +x3(y1-y2)]

Area of a Triangle
XX’
Y’
O
Y A(x1, y1)
C(x3, y3)
B(x2,y2)
M L N
Area of  ABC =
Area of trapezium ABML + Area of trapezium ALNC
- Area of trapezium BMNC

Area of a Triangle
XX’
Y’
O
Y
A(x1, y1)
C(x3, y3)
B(x2,y2)
M L N
Area of trapezium ABML + Area of trapezium ALNC
- Area of trapezium BMNC
        
1 1 1
BM AL ML AL CN LN BM CN MN
2 2 2
     
        2 1 1 2 1 3 3 1 2 3 3 2
1 1 1
y y x x y y x x y y x x
2 2 2
        
Sign of Area : Points anticlockwise  +ve
Points clockwise  -ve

• WHEN DURING COLLINEARITY OF 3 POINTS
• FORMULA Three points (x1,y1),(x2,y2),(x3,y3)
are collinear if,
[x1 ( y2 - y3) + x2( y3- y1) + x3 (y1 - y2 )] = 0

QUESTIONS
1. Find the area of ∆ shown in graph

2. Find the area of the ∆ formed by joining the mid
points of the sides of the triangle whose vertices are A
(0,-1),B (2,1) and C (0,3).Find the ratio of ar. Of ∆ formed
to the given ∆. A(0,-1)
F(1,0) E(0,1)
B(2,1) C(0,3)
D(1,2)

SOLUTION
Area of ∆ ABC ,
= 12 [x1 ( y2 - y3) + x2( y3- y1) + x3 (y1 - y2 )]
= 12 [0 (1- 3 )+2 (3+1) +0 (0-1)
= 12 [0+8+0] = 4 sq. units.
Area of ∆ DEF,
= 12 [x1 ( y2 - y3) + x2( y3- y1) + x3 (y1 - y2 )]
= 12[ 1(1-0) + 0(0-2) + 1(2-1)
= 12 [1+1] = 1sq. Units.

Maths ar. of triangles shubham n group

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