Find the area of the $MNRK$ trapezoid with the lateral side $RK = 3$ if the distances from the vertices $M$ and $N$ to the line $RK$ are $5$ and $7$, respectively.
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Tags: geometry, trapezoid, area
vanstraelen
21.06.2020 22:27
parmenides51 wrote: Find the area of the $MNRK$ trapezoid with the lateral side $PK = 3$ if the distances from the vertices $M$ and $N$ to the line $RK$ are $5$ and $7$, respectively. Is the lateral side $\ RK\ $?
AwesomeLife_Math
21.06.2020 23:34
Here's what I'm thinking.
Let the altitude dropped from $N$ to $RK$ intersect the line at $A$. Likewise, denote the same for point $M$ as $B$.
Now since $\triangle NAK$ ~ $\triangle MBK$ we can set $NK=7k$ and $MK=5k$.
Continuing on, we can drop the height of the trapezoid from $R$ onto side $MK$ (and denote the intersection as $C$).
We know that $\triangle KRC$ ~ $\triangle KMB$ as well.
Thus we get the equality $\frac{3}{h}=\frac{5k}{5}$ and $h=\frac{3}{k}$.
Finally, we can get the area of the quadrilateral by $\frac{7k+5k}{2} \cdot \frac{3}{k}=18$