SOLUTION: The base of a triangular pyramid has side lengths 10, 24, and 26. The altitude of the pyramid is 20. Find the area of a cross-section parallel to the base whose distance from the

Algebra ->  Bodies-in-space -> SOLUTION: The base of a triangular pyramid has side lengths 10, 24, and 26. The altitude of the pyramid is 20. Find the area of a cross-section parallel to the base whose distance from the      Log On


   

Question 1158313: The base of a triangular pyramid has side lengths 10, 24, and 26. The altitude of the pyramid is 20. Find the area of a cross-section parallel to the base whose distance from the base is 15.
Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The side lengths of the base make it a right triangle, with legs 10 and 24. The area of that base is (1/2)(10)(24) = 120.

The cross section is 15/20 = 3/4 of the way from the base to the peak. The cross section is a right triangle similar to the base; its side lengths are 1/4 of the corresponding side lengths of the base.

So the legs of the cross section are 5/2 and 6; the area is (1/2)(5/2)(6) = 15/2.

Notice the area of the cross section can be easily determined without finding the side lengths of the sides. The ratio of similarity between the cross section and the base is 1:4, so the ratio of their areas is 1:16. So the area of the cross section is (1/16)*120 = 120/16 = 15/2.

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

We want the area of ΔDEF

PB = 20 is the altitude, EB = 15, so PE = 20-15 = 5

ΔABC is a right triangle because

AB²+BC² = 10²+24² = 100+576 = 676 = 26² = AC²

ΔDEF ∽ ΔABC, so ΔDEF is also a right triangle.

 ΔPED ∽ ΔPBA      ΔPEF ∽ ΔPBC
PE/PB = DE/AB     PE/PB = EF/BC
 5/20 = DE/10      5/20 = EF/24
20∙DE = 5∙10      20∙EF = 5∙24
20∙DE = 50        20∙EF = 120   
   DE = 2.5          EF = 6

Area of ΔDEF = 0.5(DE)(EF) = 0.5(2.5)(6) = 7.5  <--Answer

Edwin