Answer
From the figure of the tetrahedron above, the area of the cross section whose distance from the base is 15 is ΔDEF
/PB/ is the altitude of the tetrahedron = 20
/EB/ is the distance from cross section to the base of the tretrahedron = 15
⇒/PE/ = 20 - 15 = 5
Since 26² = 10² + 24², That is
676 = 100 + 576, ΔABC is a right triangle.
Since ΔDEF ≅ ΔABC, ΔDEF is also a right triangle.
ΔPED ≅ ΔPBA
To find /DE/;
[tex]\begin{gathered} \frac{PE}{PB}=\frac{DE}{AB} \\ \frac{5}{20}=\frac{DE}{10} \\ DE=\frac{5\times10}{20} \\ DE=2.5 \end{gathered}[/tex]Since ΔPEF ≅ ΔPBC
Also, to find /EF/;
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