Q26 of 66 Page 7

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are A(2,1), B(4,3), C(2,5).(CBSE 2011)


The triangle has vertices A(2,1), B(4,3), C(2,5).


Now, mid-point of AB = D = = (3,2)


Mid-point of BC = F = = (3,4)


Mid-point of CA = E = = (2,3)


So, the triangle formed by joining the mid-points of the sides of the triangle ABC has vertices (3, 2), (3, 4) and (2, 3).


Area of triangle ABC =


1/2 |[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]|


= 1/2 |[3(3 - 4) + 3(3 - 2) + 2(2 - 4)


= 1/2 |[-3 + 3 -4]|


= 2/2


= 1 sq. units

More from this chapter

All 66 →
24

Point P(x, 4) lies on the line segment joining the points A(-5, 8) and B(4, -10). Find the ratio in which point P divides the line segment AB. Also, find the value of x.(CBSE 2011)

 25

Find the area of the quadrilateral ABCD, whose vertices are A(-3,-1), B(-2,-4), C(4,-1), and D(3,4).(CBSE 2011, 12)

 27

If (3, 3), (6, y), (x, 7) and (5, 6) are the vertices of a parallelogram taken in order, find the values of x and y.(CBSE 2011)

 28

If two vertices of an equilateral triangle are (3, 0) and (6, 0), find the third vertex.

Or


Find the value of k, if the points P(5, 4), Q(7, k) and R(9, -2) are collinear.(CBSE 2011)