Chapter 11.2, Problem 28PPS

(a)

To determine

To find: The area of the grain.

Answer to Problem 28PPS

The area of the grain is $22{\text{microns}}^2$ .

Explanation of Solution

Given:

The grain has the height of 4 microns and bases have the length of 5 and 6 microns.

Calculation:

The grain is in the shape of the trapezoid as,

  $\begin{array}{l}{b}_1=5\text{microns}\\ b_2=6\text{microns}\\ h=4\text{microns}\end{array}$

Then, the area of the trapezoid is,

  $\begin{array}{c}A=\frac{1}{2}\left(b_1+b_2\right)h\\ =\frac{1}{2}\left(5+6\right)4\\ =\frac{1}{2}\left(11+4\right)\\ =22{\text{microns}}^2\end{array}$

(b)

To determine

To find: The area of the grain.

Answer to Problem 28PPS

The area of the grain is $9.3{\text{microns}}^2$ .

Explanation of Solution

Given:

The grain has the diagonals of 3.8 microns and 4.9 microns.

Calculation:

The grain is in the shape of the rhombus as,

  $\begin{array}{l}{d}_1=3.8\text{microns}\\ d_2=4.9\text{microns}\end{array}$

Then, the area of the trapezoid is,

  $\begin{array}{c}A=\frac{1}{2}\left(d_1{d}_2\right)h\\ =\frac{1}{2}\left(3.8\right)\left(4.9\right)\\ =\frac{1}{2}\left(18.62\right)\\ =9.3{\text{microns}}^2\end{array}$