4. The legs of the trapezoid are $39 \mathrm{~mm}$ and $45 \mathrm{~mm}$, and the diagonal which is perpendicular to the longer leg has a length of $60 \mathrm{~mm}$. Construct this trapezoid, and then calculate its perimeter and area.
Solution. First, we will calculate the area and perimeter of the trapezoid. From the right-angled $\triangle ABC$ for the base $AB = a$, we have $a^2 = 60^2 + 45^2$, which means $a = 75 \text{ mm}$. Further, for the height $h$ of the trapezoid, we have $\frac{75h}{2} = \frac{60 \cdot 45}{2}$, i.e.,

$h = 36 \text{ mm}$. Now, for the projections of the legs of the base $AB$, we have $x^2 = c^2 - h^2$ and $y^2 = d^2 - h^2$, which means $x = 27 \text{ mm}$ and $y = 15 \text{ mm}$. Therefore, $b = a - (x + y) = 33 \text{ mm}$. Thus, the perimeter of the trapezoid is

$$
O = a + b + c + d = 192 \text{ mm}
$$

and its area is

$$
P = \frac{a + b}{2} h = 1944 \text{ mm}^2
$$

Construction: 1) Construct the right-angled triangle $ABC$ (the legs are given).

2) Draw a line $p \parallel AB$ through point $C$.
3) From point $A$, draw a circular arc with radius $r = 39 \text{ mm}$ and determine the point $D$ at the intersection

![](https://cdn.mathpix.com/cropped/2024_06_05_6a3de2303cb9ec4e16a2g-3.jpg?height=426&width=630&top_left_y=455&top_left_x=798)
with the line $p$.

The trapezoid $ABCD$ is the desired trapezoid.