Advertisements
Advertisements
प्रश्न
Find the area of the following triangle:

Advertisements
उत्तर
Area of triangle = `1/2 xx "Base" xx "Height"`
Base = 5 cm,
Height = 3.2 cm
Area = `1/2 xx 5 xx 3.2`
= 8 cm2
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.

Find the area of the triangle ABC with A(1, −4) and mid-points of sides through A being (2, −1) and (0, −1).
Find the area of the triangle PQR with Q(3,2) and the mid-points of the sides through Q being (2,−1) and (1,2).
The vertices of ΔABC are (−2, 1), (5, 4) and (2, −3) respectively. Find the area of the triangle and the length of the altitude through A.
Four points A (6, 3), B (−3, 5), C(4, −2) and D (x, 3x) are given in such a way that `(ΔDBG) /(ΔABG)=1/2,` find x
Find the area of ΔABC whose vertices are:
A(-5,7) , B (-4,-5) and C (4,5)
Show that the following points are collinear:
A(5,1), B(1, -1) and C(11, 4)
In a triangle ABC, if `|(1, 1, 1),(1 + sin"A", 1 + sin"B", 1 + sin"C"),(sin"A" + sin^2"A", sin"B" + sin^2"B", sin"C" + sin^2"C")|` = 0, then prove that ∆ABC is an isoceles triangle.
The value of the determinant `abs((1,"x","x"^3),(1,"y","y"^3),(1,"z","z"^3))` is ____________.
Points A(3, 1), B(12, –2) and C(0, 2) cannot be the vertices of a triangle.
