Advertisements
Advertisements
प्रश्न
A(6,1) , B(8,2) and C(9,4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ΔADE
उत्तर
Let (x,y) be the coordinates of D and ( x ',y')be thee coordinates of E. since, the diagonals of a parallelogram bisect each other at the same point, therefore
`(x+8)/2=(6+9)/2⇒x =7`
`(y+2)/2 = (1+4)/2 ⇒ y=3`
Thus, the coordinates of D are ( 7,3)
E is the midpoint of DC, therefore
`x' = (7+9)/2 ⇒ x' = 8`
`y' =(3+4)/2 ⇒ y' = 7/2 `
Thus, the coordinates of E are `(8,7/2)`
`"let" A (x_1, y_1) = A(6,1) ,E (x_2,y_2) = E (8,7/2) and D(x_3,y_3) = D(7,3) Now`
`"Area" (Δ ABC) = 1/2 [ x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)]`
`=1/2[6(7/2-3)+8(3-1)+7(1-7/2)]`
`=1/2[3/2]`
`=3/4` sq. units
Hence, the area of the triangle ΔADE is `3/ 4`sq. units
APPEARS IN
संबंधित प्रश्न
The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle.
Prove that the points (a, b + c), (b, c + a) and (c, a + b) are collinear
If `a≠ b ≠ c`, prove that the points (a, a2), (b, b2), (c, c2) can never be collinear.
Show that the points A(-5,6), B(3,0) and C(9,8) are the vertices of an isosceles right-angled triangle. Calculate its area.
For what value of k(k>0) is the area of the triangle with vertices (-2, 5), (k, -4) and (2k+1, 10) equal to 53 square units?
Using integration, find the area of the triangle whose vertices are (2, 3), (3, 5) and (4, 4).
The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is ______.
A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ∆ADE.
Ratio of the area of ∆WXY to the area of ∆WZY is 3:4 in the given figure. If the area of ∆WXZ is 56 cm2 and WY = 8 cm, find the lengths of XY and YZ.
Find the missing value:
| Base | Height | Area of Triangle |
| ______ | 31.4 mm | 1256 mm2 |
