Some Geo Problem

Geometry Level pending

In trapezoid $ABCD$ , $AD$ is parallel to $BC$ . $\angle A = \angle D = 45^{\circ}$ , while $\angle B = \angle C = 135^{\circ}$ . If $AB=6$ and the area of $ABCD$ is $30$ , find $BC$ .

2

2\sqrt{3}

2\sqrt{5}

2\sqrt{2}

1 solution

Tom Engelsman
Jan 4, 2021

Trapezoid $ABCD$ is isosceles with lengths $AB = CD = 6$ , $BC = x$ , and $AD = x + 2 \cdot 6\cos (\pi/4)$ . Its height, $h,$ is equal to $6\sin(\pi/4).$ If the area is equal to $30$ , then we can solve for $BC$ according to:

$30 = \frac{h}{2} \cdot (AD + BC) = \frac{1}{2}(6\sin(\pi/4)) \cdot [2x + 2(6\cos(\pi/4)] = (\frac{6}{\sqrt{2}})(x + \frac{6}{\sqrt{2}})$ ;

or $x = 5\sqrt{2} - 3\sqrt{2} = \boxed{2\sqrt{2}}.$

Some Geo Problem

1 solution

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