Advertisements
Advertisements
प्रश्न
Find the area of quadrilateral ABCD whose vertices are A(-5, 7), B(-4, -5) C(-1,-6) and D(4,5)
Advertisements
उत्तर
By joining A and C, we get two triangles ABC and ACD .
`" let" A (x_1,y_1)=A(-5,7) , B(x_2,y_2) = B(-4,-5) , C (x_3,y_3) = c (-1,-6) and D(x_4,y_4) = D(4,5)`
Then
`"Area of" Δ ABC = 1/2 [ x_1 (y_2 -y_3) +x_2 (y_3-y_1) +x_3(y_1-y_2)]`
`=1/2[-5(-5+6)-4(-6-7)-1(7+5)]`
`=1/2[-5+52-12]=35/2` sq. units
`"Area of" Δ ACD = 1/2 [x_1(y_3-y_4)+x_3(y_4-y_1)+x_4(y_1-y_3)]`
`=1/2 [-5(-6-5)-1(5-7)+4(7+6)]`
`=1/2[55+2+52]=109/2 `sq. units
So, the area of the quadrilateral ABCD is `35/2+109/2=72 ` sq .units.
APPEARS IN
संबंधित प्रश्न
The base PQ of two equilateral triangles PQR and PQR' with side 2a lies along y-axis such that the mid-point of PQ is at the origin. Find the coordinates of the vertices R and R' of the triangles.
Show that the points (−3, 2), (−5,−5), (2, −3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus.
Determine the ratio in which the point P (m, 6) divides the join of A(−4, 3) and B(2, 8). Also, find the value of m.
The line joining the points (2, 1) and (5, −8) is trisected at the points P and Q. If point P lies on the line 2x − y + k = 0. Find the value of k.
Find the points on the x-axis, each of which is at a distance of 10 units from the point A(11, –8).
ABCD is a rectangle whose three vertices are A(4,0), C(4,3) and D(0,3). Find the length of one its diagonal.
If A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area.
Find the value of a for which the area of the triangle formed by the points A(a, 2a), B(−2, 6) and C(3, 1) is 10 square units.
If the points A(−2, 1), B(a, b) and C(4, −1) ae collinear and a − b = 1, find the values of aand b.
Write the distance between the points A (10 cos θ, 0) and B (0, 10 sin θ).
What is the distance between the points (5 sin 60°, 0) and (0, 5 sin 30°)?
Find the area of triangle with vertices ( a, b+c) , (b, c+a) and (c, a+b).
If the points A (1,2) , O (0,0) and C (a,b) are collinear , then find a : b.
If points (t, 2t), (−2, 6) and (3, 1) are collinear, then t =
If points (a, 0), (0, b) and (1, 1) are collinear, then \[\frac{1}{a} + \frac{1}{b} =\]
If segment AB is parallel Y-axis and coordinates of A are (1, 3), then the coordinates of B are ______
Point (–3, 5) lies in the ______.
Abscissa of all the points on the x-axis is ______.
Ordinate of all points on the x-axis is ______.
The coordinates of two points are P(4, 5) and Q(–1, 6). Find the difference between their abscissas.
