9.5 Area and volume of geometric figures and objects

Find the area of the triangle.

$\begin{array}{ccc}\hfill A_T& =& \frac{1}{2}\cdot b\cdot h\hfill \\ & =& \frac{1}{2}\cdot \text{20}\cdot 6\text{ sq ft}\\ & =& \text{10}\cdot 6\text{ sq ft}\hfill \\ & =& \text{60 sq ft}\hfill \\ & =& {\text{60 ft}}^2\hfill \end{array}$

The area of this triangle is 60 sq ft, which is often written as 60 ft ² .

Find the area of the rectangle.

Let's first convert 4 ft 2 in. to inches. Since we wish to convert to inches, we'll use the unit fraction $\frac{\text{12 in}\text{.}}{\text{1 ft}}$ since it has inches in the numerator. Then,

$\begin{array}{ccc}\hfill 4\text{ft}& =& \frac{\text{4 ft}}{1}\cdot \frac{\text{12}\text{in}\text{.}}{\text{1 ft}}\hfill \\ & =& \frac{4\cancel{\text{ft}}}{1}\cdot \frac{\text{12}\text{in}\text{.}}{1\cancel{\text{ft}}}\hfill \\ & =& \text{48}\text{in}\text{.}\hfill \end{array}$

Thus, $\text{4 ft 2 in}\text{.}=\text{48 in}\text{.}+\text{2 in}\text{.}=\text{50 in}\text{.}$

$\begin{array}{ccc}\hfill A_R& =& l\cdot w\hfill \\ & =& \text{50}\text{ in}\text{.}\cdot \text{8 in}\text{.}\hfill \\ & =& \text{400 sq in}\text{.}\hfill \end{array}$

The area of this rectangle is 400 sq in.

Find the area of the parallelogram.

$\begin{array}{ccc}\hfill A_P& =& b\cdot h\hfill \\ & =& \text{10}\text{.}3\text{ cm}\cdot 6\text{.}\text{2 cm}\hfill \\ & =& \text{63}\text{.}\text{86 sq cm}\hfill \end{array}$

The area of this parallelogram is 63.86 sq cm.

Find the area of the trapezoid.

$\begin{array}{ccc}\hfill A_{\mathit{\text{Trap}}}& =& \frac{1}{2}\cdot \left(b_1+b_2\right)\cdot h\hfill \\ & =& \frac{1}{2}\cdot \left(\text{14.5 mm},+,\text{20.4 mm}\right)\cdot \left(\text{4.1 mm}\right)\hfill \\ & =& \frac{1}{2}\cdot \left(\text{34.9 mm}\right)\cdot \left(\text{4.1 mm}\right)\hfill \\ & =& \frac{1}{2}\cdot \left(\text{143.09 sq mm}\right)\hfill \\ & =& \text{71.545 sq mm}\hfill \end{array}$

The area of this trapezoid is 71.545 sq mm.

Find the approximate area of the circle.

$\begin{array}{ccc}\hfill A_c& =& \pi \cdot r^2\hfill \\ & \approx & (3.14)\cdot {\left(\text{16.8 ft}\right)}^2\hfill \\ & \approx & \left(3.14\right)\cdot \left(\text{282.24 sq ft}\right)\hfill \\ & \approx & \text{888.23 sq ft}\hfill \end{array}$

The area of this circle is approximately 886.23 sq ft.

Find the volume of the rectangular solid.

$\begin{array}{ccc}\hfill V_R& =& l\cdot w\cdot h\hfill \\ & =& \text{9 in.}\cdot \text{10 in.}\cdot \text{3 in.}\hfill \\ & =& \text{270 cu in.}\hfill \\ & =& {\text{270 in.}}^3\hfill \end{array}$

The volume of this rectangular solid is 270 cu in.

Find the approximate volume of the sphere.

$\begin{array}{ccc}\hfill V_S& =& \frac{4}{3}\cdot \pi \cdot r^3\hfill \\ & \approx & \left(\frac{4}{3}\right)\cdot \left(3.14\right)\cdot {\left(\text{6 cm}\right)}^3\hfill \\ & \approx & \left(\frac{4}{3}\right)\cdot \left(3.14\right)\cdot \left(\text{216 cu cm}\right)\hfill \\ & \approx & \text{904.32 cu cm}\hfill \end{array}$

The approximate volume of this sphere is 904.32 cu cm, which is often written as 904.32 cm ³ .

Find the approximate volume of the cylinder.

$\begin{array}{ccc}\hfill V_{\mathrm{Cyl}}& =& \pi \cdot r^2\cdot h\hfill \\ & \approx & \left(3.14\right)\cdot {\left(\text{4.9 ft}\right)}^2\cdot \left(\text{7.8 ft}\right)\hfill \\ & \approx & \left(3.14\right)\cdot \left(\text{24.01 sq ft}\right)\cdot \left(\text{7.8 ft}\right)\hfill \\ & \approx & \left(3.14\right)\cdot \left(\text{187.278 cu ft}\right)\hfill \\ & \approx & \text{588.05292 cu ft}\hfill \end{array}$

The volume of this cylinder is approximately 588.05292 cu ft. The volume is approximate because we approximated $\pi$ with 3.14.

Find the approximate volume of the cone. Round to two decimal places.

$\begin{array}{ccc}\hfill V_c& =& \frac{1}{3}\cdot \pi \cdot r^2\cdot h\hfill \\ & \approx & \left(\frac{1}{3}\right)\cdot \left(3.14\right)\cdot {\left(\text{2 mm}\right)}^2\cdot \left(\text{5 mm}\right)\hfill \\ & \approx & \left(\frac{1}{3}\right)\cdot \left(3.14\right)\cdot \left(\text{4 sq mm}\right)\cdot \left(\text{5 mm}\right)\hfill \\ & \approx & \left(\frac{1}{3}\right)\cdot \left(3.14\right)\cdot \left(\text{20 cu mm}\right)\hfill \\ & \approx & \text{20}\text{.}9\overline{3}\text{ cu mm}\hfill \\ & \approx & \text{20}\text{.}\text{93 cu mm}\hfill \end{array}$

The volume of this cone is approximately 20.93 cu mm. The volume is approximate because we approximated $\pi$ with 3.14.