Question

A trapezoid has vertices at A(5, 0)A(5, 0) , B(4, 3)B(4, 3) , C(10, 5)C(10, 5) , and D(8, 1)D(8, 1) , and angles A and B are right angles. What is the area of the trapezoid

Answer 1

check the picture below.

$\bf \textit{area of this trapezoid}\\\\ A=\cfrac{{AB}({BC}+{AD})}{2}\\\\ -------------------------------\\\\ \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) A&({{ 5}}\quad ,&{{ 0}})\quad % (c,d) B&({{ 4}}\quad ,&{{ 3}})\\ B&({{ 4}}\quad ,&{{ 3}})\quad % (c,d) C&({{ 10}}\quad ,&{{ 5}})\\ A&({{ 5}}\quad ,&{{ 0}})\quad % (c,d) D&({{ 8}}\quad ,&{{ 1}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}$

$\bf AB=\sqrt{(4-5)^2+(3-0)^2}\implies AB=\sqrt{1+9}\implies AB=\sqrt{10} \\\\\\ BC=\sqrt{(10-4)^2+(5-3)^2}\implies BC=\sqrt{36+4}\implies BC=\sqrt{40} \\\\\\ AD=\sqrt{(8-5)^2+(1-0)^2}\implies AD=\sqrt{9+1}\implies AD=\sqrt{10}\\\\ -------------------------------$

$\bf \begin{cases} \sqrt{40}\\ \qquad \sqrt{2^2\cdot 10}\\ \qquad 2\sqrt{10} \end{cases}\qquad A=\cfrac{h(x+y)}{2}\implies A=\cfrac{\sqrt{10}(2\sqrt{10}+\sqrt{10})}{2} \\\\\\ A=\cfrac{\sqrt{10}(3\sqrt{10})}{2}\implies A=\cfrac{3(\sqrt{10})^2}{2}\implies A=\cfrac{30}{2}\implies A=15$