3. A trapezoid $A B C D$ is given, in which $|A B|=8 \mathrm{~cm}$ and $|\angle A B C|=90^{\circ}$. Its perimeter is $28 \mathrm{~cm}$. A semicircle $k$ with diameter $A B$ touches side $C D$. Calculate the lengths of the remaining sides of the given trapezoid, if side $A B$ is its

a) base,

b) leg.
3. Let $S$ be the midpoint of side $AB$ and $T$ the point of tangency of the semicircle $k$ with side $CD$. If $AB$ is the base of the given trapezoid, then $CD \parallel AB$ and $|AB| + |BC| + |CT| = 2|AB| = 16 \, \text{cm}$ (Fig. 2). Let $A_1$ be the orthogonal projection of vertex $A$ onto line $CD$. Since $|TD| + |DA| = 28 \, \text{cm} - 16 \, \text{cm} = 12 \, \text{cm} > |AA_1| + |A_1T| = 8 \, \text{cm}$, vertex $D$ lies on the ray $TA_1$ beyond point $A_1$. Let the length $|A_1D| = x \, \text{cm}$ and $|DA| = d \, \text{cm}$. For the numbers $x, d$, we get the system of equations $d + x = 8, d^2 = x^2 + 4^2$ (Pythagoras' theorem for triangle $AA_1D$), which can be easily transformed into the form $d + x = 8, (d - x)(d + x) = 16$, i.e., $d + x = 8, d - x = 2$. The system has a unique solution $d = 5, x = 3$. The remaining sides of the given trapezoid thus have lengths $4 \, \text{cm}, 11 \, \text{cm}$, and $5 \, \text{cm}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_fa4e6f4674eb91bf6d8dg-3.jpg?height=303&width=596&top_left_y=1087&top_left_x=426)

Fig. 2

![](https://cdn.mathpix.com/cropped/2024_04_17_fa4e6f4674eb91bf6d8dg-3.jpg?height=507&width=591&top_left_y=896&top_left_x=1168)

Fig. 3

If $AB$ is a leg of the given trapezoid $ABCD$, then $AB \perp BC \parallel AD$ (Fig. 3), so both bases $BC$ and $AD$ touch the semicircle $k$ at the corresponding vertices $B$ and $A$. Let $b$ be the equal segments of the tangents from vertex $C$ and $d$ the equal segments of the tangents from vertex $D$ to the semicircle $k$. From the knowledge of the perimeter, we get (in centimeters) the equality $28 = 8 + 2b + 2d$, i.e., $b + d = 10$. From the parallelism $BC \parallel AD$, it follows that $|b - d| = \sqrt{(b + d)^2 - 8^2} = 6$. Given the symmetry of $k$ with respect to the axis of the semicircle $k$, it suffices to consider only one of the possibilities, for example, $b > d$. The system $b + d = 10, b - d = 6$ has a unique solution $b = 8, d = 2$, so the remaining sides of the given trapezoid have lengths $8 \, \text{cm}, 10 \, \text{cm}$, and $2 \, \text{cm}$, which is also true in the case $b < d$.

Full solution is worth 6 points, each correct solution is worth 3 points.