Advertisements
Advertisements
प्रश्न
Find the area of ΔABC whose vertices are:
A(10,-6) , B (2,5) and C(-1,-3)
उत्तर
A(10,-6) , B (2,5) and C(-1,-3) are the vertex of ΔABC. Then,
`(x_1=10,y_1=-6),(x_2=2,y_2=5) and (x_3=-1,y_3=3)`
Area of triangle ABC
`=1/2{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}`
`=1/2{10(5-3)+2(3-(-6))+(-1)(-6-5)}`
`=1/2{10(2)+2(9)-1(-11)}`
`=1/2{20+18+11}`
`=1/2(49)`
=24.5 sq. units
APPEARS IN
संबंधित प्रश्न
The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the following figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of Δ PQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?
Find the area of a triangle with vertices at the point given in the following:
(−2, −3), (3, 2), (−1, −8)
Find the area of the following triangle:
The point A divides the join of P (−5, 1) and Q(3, 5) in the ratio k:1. Find the two values of k for which the area of ΔABC where B is (1, 5) and C(7, −2) is equal to 2 units.
Show that the points (-3, -3),(3,3) and C (-3 `sqrt(3) , 3 sqrt(3))` are the vertices of an equilateral triangle.
A(6,1) , B(8,2) and C(9,4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ΔADE
Find the value of m if the points (5, 1), (–2, –3) and (8, 2m) are collinear.
Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.
Find the missing value:
| Base | Height | Area of parallelogram |
| ______ | 8.4 cm | 48.72 cm2 |
Area of triangle MNO in the figure is ______.
