Answer to Question #150267 in Geometry for Asfand Khan

Question

At the base of the triangular pyramid DABC lies an isosceles acute-angled triangle ABC (AC = BC). It is known that CB > AD, and the edge DA is perpendicular to the plane ABC. The projections of the pyramid are considered DABC on the planes containing the straight line AC. It is known that the largest area of such a projection is 37, the smallest is 12, and the area of the triangle ABC is 35. Find the volume of the pyramid DABC. In response, write down the square of the volume.

Answer 1

According to the figure above, the largest triangle would be ∆BCD and the smallest ∆ABD and ∆ACD which are equal to each other

Therefore, Area Isosceles ∆ABC = 35

Area ∆BCD = 37

Area ∆ABD = 12

Area ∆ACD = 12

volume of a triangular pyramid = 1/3 × base area × h

= 1/3 × Area ∆ABC × |AD|

for ∆ABD, 1/2bh = 12

1/2 × |AD| × |AB| = 12

|AD| × |AB| = 24

Since |AD| and |AB| are part of a Pythagoras Triple as

|AD|² + |AB|² = |BD|²

|AD| = 3; |AB| = 8

but, volume of a triangular pyramid = 1/3 × Area ∆ABC × |AD| = 1/3 × 35 × 3 = 35

square of the volume = 35² = 1225