Question

Given: ABCD is a trapezoid, AC ⊥ CD AB = CD, AC=the square root of 75 , AB = 5 Find: AABCD

Answer 1

In this attached picture according to the conditions of the problem we have an isosceles trapezoid and since we know that legs are equal (AD=BC=5 cm), we have to calculate bases and height in order to find the area. Working with the triangle BCD, we apply Pythagoras theorem and find that CD = $\sqrt{75+25}$ = 10 cm. Since BDC is a right triangle, applying theorem for the area of triangles, we find that $\frac{1}{2} * BF = \frac{1}{2} * 5 * \sqrt{75}$ and BF= $0.5\sqrt{75}$. Since ABCD is an isosceles trapezoid, triangles ADE and BFC are congruent with Angle Side Angle theorem. Then, DE=FC and with the help of Pythagoras theorem, DE=FC=2.5 cm. Then, AB=EF=5 cm and the area of the trapezoid is  $A= BF * \frac{AB+CD}{2} = 0.5 \sqrt{75} * \frac{5+10}{2} = 18.75 \sqrt{3} cm^{2}$