Question

The base of tetrahedron is a triangle whose sides have lengths 10, 24,26 . The altitude of the tetrahedron is 20. find the area of the cross section whose distance from the base is 15.

Answer 1

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/;

`\begin{gathered} (PE)/(PB)=(DE)/(AB) \\ (5)/(20)=(DE)/(10) \\ DE=(5*10)/(20) \\ DE=2.5 \end{gathered}`

Since ΔPEF ≅ ΔPBC

Also, to find /EF/;