Question

Show that the quadrilateral ABCD with vertices A(0, 3), B(–2, 0), C(0, –5) and D(2, 0) is a kite. Also, find the length of each diagonal of the kite ABCD.

Answer 1

1. Calculate side lengths

Using the distance formula `d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2` for vertives A(0, 3), B(–2, 0), C(0, –5) and D(2, 0):

Side AB: `sqrt((-2 - 0)^2 + (0 - 3)^2`

= `sqrt(4 + 9)`

= `sqrt(13)`

Side AD: `sqrt((2 - 0)^2 + (0 - 3)^2`

= `sqrt(4 + 9)`

= `sqrt(13)`

Side BC: `sqrt((0 - (-2))^2 + (-5 - 0)^2`

= `sqrt(4 + 25)`

= `sqrt(29)`

Side CD: `sqrt((2 - 0)^2 + (0 - (-5))^2`

= `sqrt(4 + 25)`

= `sqrt(29)`

Since AB = AD and BC = CD, the quadrilateral has two pairs of equal adjacent sides, which is the definition of a kite.

2. Find lengths of diagonals

The diagonals are the segments connecting opposite vertices: AC and BD.

Diagonal AC: The distance between (0, 3) and (0, –5) is `sqrt((0 - 0)^2 + (-5 - 3)^2`

= `sqrt(64)`

= 8 units

Diagonal BD: The distance between (–2, 0) and (2, 0) is `sqrt((2 - (-2))^2 + (0 - 0)^2`

= `sqrt(16)`

= 4 units

The quadrilateral ABCD is a kite because it has two pairs of equal adjacent sides (AB = AD = `sqrt(13)` and BC = CD = `sqrt(29)`). The length of diagonal AC is 8 units and the length of diagonal BD is 4 units.