Question

3. Find the perimeter of quadrilateral ABCD with vortices 4(-2,-2), B(-1,3),
c(5, 3), and D(4, -2).

Answer 1

Answer:

$12 + 2 \sqrt{26}$        

Step-by-step explanation:

To find the perimeter of quadrilateral we need to calculate the length of its four sides which is distance between its four vertices.

where $A(x_{1}, y_{1}) = (-2, -2)$,

$B(x_{2}, y_{2}) = (-1, 3)$

$C(x_{3}, y_{3}) = (5, 3)$

$D(x_{4}, y_{4}) = (4, -2)$

Distance of AB = $\sqrt{\left (x^{_{2}}-x^{_{1}} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$

= $\sqrt{\left (-1+ 2 \right )^{2}+ (3 + 2)^{^{2}}}$

= $\sqrt{1^{^{2}}+5^{^{2}}}$

=$\sqrt{26}$

Distance of BC = $\sqrt{\left (x_{3}-x_{2} \right )^{2}+ \left (y_{3}-y_{2}\right )^{2}}$

= $\sqrt{(5+1)^{2}+ (3-3)^{2}} = \sqrt{36} = 6$

Distance of CD = $\sqrt{\left (x_{4}-x_{3} \right )^{2}+ \left (y_{4}-y_{3}\right )^{2}$

= $\sqrt{(4-5)^{^{2}}+(-2-3)^{^{2}}} = \sqrt{1 + 25} =\sqrt{26}$

Distance of AD = $\sqrt{\left (x_{4}-x_{1} \right )^{2}+ \left (y_{4}-y_{1} \right )^{2}$

= $\sqrt{(4 + 2)^{_{2}}+ (-2 + 2)^{^{2}}}=\sqrt{36 + 0} = 6$

Perimeter of quadrilateral ABCD = Dist. ( AB + BC + CD + AD)

= $\sqrt{26} +6 +\sqrt{26} +6$

=$12 +2 \sqrt{26}$