[ Area of a trapezoid
[ Projections of the bases, sides, or vertices of the trapezoid ]

The bases of an isosceles trapezoid are equal to $a$ and $b(a>b)$, and the acute angle is $45^{\circ}$. Find the area of the trapezoid.

#
The projection of the lateral side of an isosceles trapezoid onto its larger base is equal to half the difference of the bases.

## Solution

From vertex $B$ of the smaller base $BC$ of the isosceles trapezoid $ABCD$, drop a perpendicular $BH$ to its larger base $AD$. Then

$$
BH = AH = \frac{1}{2}(AD - BC) = \frac{a - b}{2}
$$

Therefore,

$$
S(ABCD) = \frac{1}{2}(AD + BC) \cdot BH = \frac{a + b}{2} \cdot \frac{a - b}{2} = \frac{a^2 - b^2}{4}
$$

## Answer

$\frac{a^2 - b^2}{4}$.