Advertisements
Advertisements
प्रश्न
Area of a triangle = `1/2` base × ______.
Advertisements
उत्तर
Area of a triangle = `1/2` base × height.
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.

In Fig. 6, ABC is a triangle coordinates of whose vertex A are (0, −1). D and E respectively are the mid-points of the sides AB and AC and their coordinates are (1, 0) and (0, 1) respectively. If F is the mid-point of BC, find the areas of ∆ABC and ∆DEF.

If the points P(–3, 9), Q(a, b) and R(4, – 5) are collinear and a + b = 1, find the values of a and b.
Find the equation of the line joining (1, 2) and (3, 6) using the determinants.
Prove that the points (2,3), (-4, -6) and (1, 3/2) do not form a triangle.
If `a≠ b ≠ c`, prove that the points (a, a2), (b, b2), (c, c2) can never be collinear.
Show that the following points are collinear:
(i) A(2,-2), B(-3, 8) and C(-1, 4)
If the points (3, -2), (x, 2), (8, 8) are collinear, then find the value of x.
If the points (2, -3), (k, -1), and (0, 4) are collinear, then find the value of 4k.
Triangles having the same base have equal area.
