Chapter 5.8, Problem 26E

(a)

To determine

To find: The area of the rhombus.

Answer to Problem 26E

  $\text{Area of rhombus}=211.2{\text{ cm}}^2$

Explanation of Solution

Given information: Geometry: The sides of a rhombus are 15 centimeters long and the length of its longer diagonal is 24.6 centimeters.

Calculation:

  

Determine the semi perimeter

  $s=\frac{1}{2}\left(15+15+24.6\right)=27.3$

Use Heron’s formula

  $\begin{array}{l}\text{triangle area}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ =\sqrt{27.3\left(27.3-15\right)\left(27.3-15\right)\left(27.3-24.6\right)}\\ \approx 105.6\end{array}$

Rhombus area = 2 (area of yellow triangle)

  $=2\times 105.6=211.2\text{}{\text{cm}}^2$

Thus,

  $\text{Area of rhombus}=211.2{\text{ cm}}^2$

(b)

To determine

To find: The measures of the angles of the rhombus.

Answer to Problem 26E

  $\begin{array}{l}A=C={110.6}^{\circ }\\ B=D={69.8}^{\circ }\end{array}$

Explanation of Solution

Given information: Geometry: The sides of a rhombus are 15 centimeters long and the length of its longer diagonal is 24.6 centimeters.

Calculation:

  

Use law of cosines

  ${24.6}^2={15}^2+{15}^2-2\times 15\times 15\cos \left(A\right)$

  $\cos \left(A\right)=\frac{{24.6}^2-\left({15}^2+{15}^2\right)}{-2\times 15\times 15}$

  $A=\text{ar}\cos \left(\frac{{24.6}^2-\left({15}^2+{15}^2\right)}{-2\times 15\times 15}\right)={110.6}^{\circ }$

Since,

C = A (rhombus) conclude that

  $C={110.6}^{\circ }$

We have

  $A+B+C+D={360}^{\circ }$

Since A = C and B = D concludes that:

  $\begin{array}{l}2A+2B={360}^{\circ }\\ \to A+B=\frac{{360}^{\circ }}{2}={180}^{\circ }\\ \to {110.6}^{\circ }+B={180}^{\circ }\\ \to B={180}^{\circ }-{110.6}^{\circ }\\ B={69.8}^{\circ }\end{array}$

Thus,

  $\begin{array}{l}A=C={110.6}^{\circ }\\ B=D={69.8}^{\circ }\end{array}$