Acat scan of a human pancreas shows cross-sections spaced 1 cm apart. the pancreas is 12 cm long and the cross-sectional areas, in square centimeters, are 0, 7.9, 15.3, 18.1, 10.2, 10.7, 9.5, 8.5, 7.8, 5.6, 4.2, 2.7, and 0. use the midpoint rule to estimate the volume of the pancreas.

100.5 cubic centimeters.

Step-by-step explanation:

The 12 intervals of the scanning are the 1-cm width intervals

[0,1), [1,2), [2,3), [3,4), [4,5), [5,6), [6,7), [7,8), [8,9), [9,10), [10,11), [11,12] and their midpoints are

0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5, 10.5 and 11.5

The idea here is to the volume of the pancreas by approximating it with the volume V of 12 cylinders of height h=1 and area of the base A equals to the area given for the scanning where that midpoints fall.

Since the volume of a cylinder equals area of the base times height and all the cylinders have the same height of 1 cm, the sum of the volumes of the 12 cylinders equals the sum of the areas of the cross-sections multiplied by 1 cm (so the result is given in cubic centimeters)

Estimation of the volume of the pancreas

7.9+15.3+18.1+10.2+10.7+9.5+8.5+7.8+5.6+4.2+2.7 cubic cm

=100.5 c.c.

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

25.45584