Question

1. Find the area of the triangle whose vertices are :
0 (2,3).(-1,0). (2.-4)
(ii) (-5, -1), (3,-5), (5,2)
(ii) (8,1),(k, -4), (2,-5)
(0) (7.-2), (5, 1). (3.)
given triangle
4. Find the area of the quadrilateral whose vertices, taken in order, are (-4,-2), (-3,-5)
(3.-2) and (2.3).
5. You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides
it into two triangles of equal areas. Verify this result for A ABC whose vertices are
A(4.-6. B(3,-2) and C(5,2).
2. In each of the following find the value of 'k', for which the points are collinear.
3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle
whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the​

Answer 1

Answer:

The area of the triangle ABC is 10.5 square unit.

Step-by-step explanation:

To find : The area of triangle ABC whose vertices are A(2,3) B(-1,0)and C(2,-4) ?

Solution :

The area of the triangle with vertices is

$A= (\frac{1}{2}) [x_1 (y_2- y_3 )+x_2 (y_3-y_1 )+x_3(y_1-y_2)]$

Here, $(x_1,y_1)=(2,3)[tex],  [tex](x_2,y_2)=(-1,0)$ and  $(x_3,y_3)=(2,-4)$.

Substitute the value,

$A= (\frac{1}{2}) [2(0-(-4) )+(-1)(-4-3)+2(3-0)]$

$A= (\frac{1}{2}) [2(4)+(-1)(-7)+2(3)]$

$A= (\frac{1}{2}) [8+7+6]$

$A= (\frac{1}{2}) [21]$

$A=\frac{21}{2}$

$A=10.5$

Therefore, the area of the triangle ABC is 10.5 square unit.