If the diagonals of an isosceles trapezoid are perpendicular to each other, prove that the area is $S=H^2$. [closed]

$\begingroup$

Where H is the altitude of the whole trapezoid (the distance between the bases). Thanks.

$\endgroup$ 4

2 Answers

$\begingroup$

Here's another way to look at it.

Let $ABCD$ be the given isosceles trapezoid with $BC=a$ the top line segment and $AD=b$ the bottom line segment (see figure). Then the area of the trapezoid is given by $$S=bH - \left( \frac{b-a}{2}\right) H = H \frac{a+b}{2}.$$ It therefore suffices to show that $H=(a+b)/2$. To do this, extend $a$ and draw the line perpendicular to $b$ through $D$ such that it intersects $a$ at $X$. Similarly, draw the line perpendicular to $b$ through $B$ intersecting $b$ at $Y$. To show that $H=(a+b)/2$ it suffices to show that the quadrilateral $BXDY$ is a square. But this is immediate, as can be seen by a translation of the diagonal $AC$ by $AY$ along $b$ (since the diagonals are of equal length, are perpendicular and span a quadrilateral, the quadrilateral must be a square). Hence $H=(a+b)/2$ and we are done.

$\endgroup$ $\begingroup$

Hint: Let $D$ be the length of a diagonal. You can readily show (by decomposing the trapezoid into two triangles) that the area of the trapezoid is $\frac12 D^2.$

Now show that $H=\frac{\sqrt2}2D.$

$\endgroup$ 1