Question

show that the points (-4,-7),(5,10),(15,8) and (3,-5) taken in order are the vertices of a rhombhus. and find its area

Answer 1

hello.....

let's Consider the points
(-7, -3), (5, 10), (15, 8) and (3, -5).

Distance between
(-7, -3), (5, 10) = √[(5 + 7)2 + (10 + 3)2] = √(144 + 169) = √313

Distance between
(5, 10), (15, 8) = √[(15 - 5)2 + (8 - 10)2] = √(100 + 4) = √104

Distance between
(15, 8), (3, -5) = √[(3 - 15)2 + (-5 - 8)2] = √(144 + 169) = √313

Distance between
(3, -5), (-7, -3) = √[(-7 - 3)2 + (-3 + 5)2] = √(100 + 4) = √104

Opposite sides of the quadrilateral formed by the given points are equal.
So, it's parallelogram.

Diagonal of a parallelogram divides it into two congruent triangles.

So, area parallelogram = 2[Area of the triangle formed by (-7, -3), (5, 10) and (15, 8).

= 2[1/2 |-7(10 - 8) + 5(8 + 3) + 15(-3 - 10)|]

= |-7(2) + 5(11) + 15(-13)|

= |-14 + 55 - 195|

= |- 154|

= 154 sq units.



thanking you...

☺