Question

Find the exact lengths of the sides of quadrilateral ABCD with vertice
A(4, 5), B(-3, 3), C(-6, -13) and D(6, - 2).

Answer 1

Answer:

$AB=\sqrt{53}$

$BC=\sqrt{265}$

$CD=\sqrt{265}$

$DA=\sqrt{53}$

ABCD is a kite.

Step-by-step explanation:

Given vertices of quadrilateral ABCD:

• A = (4, 5)
• B = (-3, 3)
• C = (-6, -13)
• D = (6, -2)

To find the exact lengths of the sides of the quadrilateral, use the distance formula.

$\boxed{\begin{minipage}{7.4 cm}\underline{Distance between two points}\\\\ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\\\\where (x_1,y_1) and (x_2,y_2) are the two points.\\\end{minipage}}$

$\begin{aligned}AB&=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\&=\sqrt{(-3-4)^2+(3-5)^2}\\&=\sqrt{(-7)^2+(-2)^2}\\&=\sqrt{49+4}\\&=\sqrt{53}\\\end{aligned}$

$\begin{aligned}BC&=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}\\&=\sqrt{(-6-(-3))^2+(-13-3)^2}\\&=\sqrt{(-3)^2+(-16)^2}\\&=\sqrt{9+256}\\&=\sqrt{265}\\\end{aligned}$

$\begin{aligned}CD&=\sqrt{(x_D-x_C)^2+(y_D-y_C)^2}\\&=\sqrt{(6-(-6))^2+(-2-(-13))^2}\\&=\sqrt{(12)^2+(11)^2}\\&=\sqrt{144+121}\\&=\sqrt{265}\\\end{aligned}$

$\begin{aligned}DA&=\sqrt{(x_A-x_D)^2+(y_A-y_D)^2}\\&=\sqrt{(4-6)^2+(5-(-2))^2}\\&=\sqrt{(-2)^2+(7)^2}\\&=\sqrt{4+49}\\&=\sqrt{53}\\\end{aligned}$

A kite has two pairs of adjacent equal sides.  

Therefore, quadrilateral ABCD is a kite as:

• AB is adjacent to DA and AB = DA.
• BC is adjacent to CD and BC = CD.