Q:

# A trapezoid has coordinates of (-5, -3), (-2, 5), (2, 5), and (5, -3). What is the approximate perimeter of the trapezoid?Round your answer to the nearest whole number (number that is not a decimal).

Accepted Solution

A:
Answer:The approximate perimeter of the trapezoid is 31 unitsStep-by-step explanation:step 1Plot the trapezoidLetA(-5, -3), B(-2, 5), C(2, 5), and D(5, -3)see the attached figurestep 2Find the perimeter of trapezoidwe know thatThe perimeter of trapezoid is equal to$$P=AB+BC+CD+AD$$the formula to calculate the distance between two points is equal to $$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$$ Find the distance ABwe have$$A(-5, -3),B(-2, 5)$$substitute in the formula$$d=\sqrt{(5+3)^{2}+(-2+5)^{2}}$$ $$d=\sqrt{(8)^{2}+(3)^{2}}$$ $$d_A_B=\sqrt{73}\ units$$ Find the distance BCwe have$$B(-2, 5),C(2, 5)$$substitute in the formula$$d=\sqrt{(5-5)^{2}+(2+2)^{2}}$$ $$d=\sqrt{(0)^{2}+(4)^{2}}$$ $$d_B_C=4\ units$$ Find the distance CDwe have$$C(2, 5),D(5, -3)$$substitute in the formula$$d=\sqrt{(-3-5)^{2}+(5-2)^{2}}$$ $$d=\sqrt{(-8)^{2}+(3)^{2}}$$ $$d_C_D=\sqrt{73}\ units$$ Find the distance ADwe have$$A(-5, -3),D(5, -3)$$substitute in the formula$$d=\sqrt{(-3+3)^{2}+(5+5)^{2}}$$ $$d=\sqrt{(0)^{2}+(10)^{2}}$$ $$d_A_D=10\ units$$ step 3Find the perimeter$$P=AB+BC+CD+AD$$substitute the values$$P=\sqrt{73}+4+\sqrt{73}+10$$$$P=31\ units$$