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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

User Bogdan B
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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=√((4-5)^2+(3-0)^2)\implies AB=√(1+9)\implies AB=√(10) \\\\\\ BC=√((10-4)^2+(5-3)^2)\implies BC=√(36+4)\implies BC=√(40) \\\\\\ AD=√((8-5)^2+(1-0)^2)\implies AD=√(9+1)\implies AD=√(10)\\\\ -------------------------------


\bf \begin{cases} √(40)\\ \qquad √(2^2\cdot 10)\\ \qquad 2√(10) \end{cases}\qquad A=\cfrac{h(x+y)}{2}\implies A=\cfrac{√(10)(2√(10)+√(10))}{2} \\\\\\ A=\cfrac{√(10)(3√(10))}{2}\implies A=\cfrac{3(√(10))^2}{2}\implies A=\cfrac{30}{2}\implies A=15
A trapezoid has vertices at A(5, 0)A(5, 0) , B(4, 3)B(4, 3) , C(10, 5)C(10, 5) , and-example-1
User Paulo Pessoa
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