In triangle $A B C$, angle $B A C$ is right, the lengths of sides $A B$ and $B C$ are 1 and 3, respectively. Point $K$ divides side $A C$ in the ratio 7:1, counting from point $A$. Which is greater: the length of $A C$ or the length of $B K$?
Calculate $B K$ and $A C$.

## Solution

$A C=\sqrt{B C^{2}-A B^{2}}=2 \sqrt{2}, A K=7 \cdot A C / 8=7 \sqrt{2} / 4$,

$$
B K=\sqrt{A K^{2}+A B^{2}}=\sqrt{49 / 8+1}=\sqrt{57 / 8}=\sqrt{114} / 4<2 \sqrt{2}
$$

(since $144<128$).

![](https://cdn.mathpix.com/cropped/2024_05_06_d7c3832c1c4b0d15493ag-25.jpg?height=343&width=824&top_left_y=2395&top_left_x=617)

## Answer

$B K<A C$.

## Problem