[ Projections of the bases, sides, or vertices of a trapezoid ]  [ Properties and characteristics of an isosceles triangle. ]

In a right trapezoid, the smaller diagonal is equal to the larger lateral side.

Find the larger diagonal if the larger lateral side is equal to $a$, and the smaller base is equal to $b$.
Prove that the larger base is twice as large as the smaller one.

## Solution

Let $CD$ be the larger lateral side of trapezoid $ABCD$, $BC$ be the smaller base, $CD=a$, $BC=b$, and $BD$ be the larger diagonal. Drop a perpendicular $CM$ to the larger base $AD$. Since $AC=CD$, then $AM=MD$, and since $AM=BC=b$, then $AD=2b$. Therefore, $AB^2=CM^2=a^2-b^2$. Consequently, $BD^2=AB^2+AD^2=4b^2+a^2-b^2=a^2+3b^2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c1015ea0b2d2805d8876g-02.jpg?height=317&width=620&top_left_y=1076&top_left_x=725)

## Answer

$\sqrt{a^2+3b^2}$.