$\left[\begin{array}{l}\text { Rhombi. Properties and characteristics } \\ {\left[\begin{array}{l}\text { Area calculation }\end{array}\right]} \\ {[\text { Thales' theorem and the theorem of proportional segments ] }}\end{array}\right]$
$[$ Law of Cosines

Given a rhombus $A B C D$ with an obtuse angle at vertex $A$. On the extension of side $A D$ beyond point $D$, a point $K$ is taken. Segments $B K$ and $C D$ intersect at point $L$.

Find the area of triangle $A B K$, if $B L=2, K L=5$, and the height of the rhombus is 1. #
Let $H$ be the projection of point $L$ onto line $AB$. Then $LH=1$. Therefore, in the right triangle $BHL$, the leg $LH$ is half the hypotenuse $BL$, so $\angle LBH=30^{\circ}$. Let $AD=2x$. Then $DK=5x$. By the cosine theorem, $AK^2=AB^2+BK^2-2AB \cdot BK \cos 30^{\circ}$, or $49x^2=4x^2+49-2 \cdot 2x \cdot 7 \cdot \frac{\sqrt{3}}{2}, 45x^2-14x\sqrt{3}-49=0$. From this, we find that $x=\frac{7\sqrt{3}}{15}$. Let $P$ be the projection of point $B$ onto line $AK$. Then $BP$ is the height of the rhombus $ABCD$ and triangle $ABK$. Therefore, $S_{ABK}=\frac{1}{2}AK \cdot BP=\frac{1}{2}7x \cdot 1=\frac{7}{2}x=\frac{7}{2} \cdot \frac{7\sqrt{3}}{15}=\frac{49\sqrt{3}}{30}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5ff42d6d57f9dccbb1f5g-49.jpg?height=688&width=1929&top_left_y=-1&top_left_x=99)

## Answer

$\frac{49\sqrt{3}}{30}$

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