Advertisements
Advertisements
प्रश्न
Find the area of the triangle whose vertices are: (2, 3), (-1, 0), (2, -4)
उत्तर
Area of a triangle is given by
Area of triangle = `1/2 {x_1 (y_2 - y_3)+ x_2 (y_3 - y_1)+ x_3 (y_1 - y_2)}`
Area of the given triangle = `1/2 [2 { 0- (-4)} + (-1) {(-4) - (3)} + 2 (3 - 0)]`
`= 1/2 {8 + 7 + 6}`
=`21/2` square units.
APPEARS IN
संबंधित प्रश्न
In Fig. 8, the vertices of ΔABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment DE is drawn to intersect the sides AB and AC at D and E respectively such that `(AD)/(AB)=(AE)/(AC)=1/3 `Calculate th area of ADE and compare it with area of ΔABCe.
Prove that the area of a triangle with vertices (t, t −2), (t + 2, t + 2) and (t + 3, t) is independent of t.
Prove that the points (2, – 2), (–3, 8) and (–1, 4) are collinear
The coordinates of A, B, C are (6, 3), (–3, 5) and (4, – 2) respectively and P is any point (x, y). Show that the ratio of the areas of triangle PBC and ABC is
The vertices of a ΔABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that `(AD)/(AB) = (AE)/(AC) = 1/4`Calculate the area of the ΔADE and compare it with the area of ΔABC. (Recall Converse of basic proportionality theorem and Theorem 6.6 related to ratio of areas of two similar triangles)
Find values of k if area of triangle is 4 square units and vertices are (k, 0), (4, 0), (0, 2)
The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle ?
Prove that the points A (a,0), B( 0,b) and C (1,1) are collinear, if `( 1/a+1/b) =1`.
The area of the triangle whose vertices are A(1, 2), B(-2, 3) and C(-3, -4) is ______.
Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9).