Advertisements
Advertisements
प्रश्न
Find the area of the triangle whose vertices are: (–5, –1), (3, –5), (5, 2)
उत्तर
Area of the given triangle = `1/2 [-5 { (-5)- (4)} + 3(2-(-1)) + 5{-1 - (-5)}]`
`= 1/2{35 + 9 + 20}`
= 32 square units
APPEARS IN
संबंधित प्रश्न
Find the values of k so that the area of the triangle with vertices (k + 1, 1), (4, -3) and (7, -k) is 6 sq. units.
If `a≠ b ≠ c`, prove that the points (a, a2), (b, b2), (c, c2) can never be collinear.
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 ?
Show that the points (-3, -3),(3,3) and C (-3 `sqrt(3) , 3 sqrt(3))` are the vertices of an equilateral triangle.
Find the third vertex of a ΔABC if two of its vertices are B(-3,1) and C (0,-2) and its centroid is at the origin
A(7, -3), B(5,3) and C(3,-1) are the vertices of a ΔABC and AD is its median. Prove that the median AD divides ΔABC into two triangles of equal areas.
For what value of x are the points A(-3, 12), B(7, 6) and C(x, 9) collinear.
If the centroid of ΔABC having vertices A (a,b) , B (b,c) and C (c,a) is the origin, then find the value of (a+b+c).
The points (0, 5), (0, –9) and (3, 6) are collinear.
In the given figure, ratio of the area of triangle ABC to the area of triangle ACD is the same as the ratio of base BC of triangle ABC to the base CD of ΔACD.
