Chapter 5.6, Problem 55E

Solution

To find: The area of the segment intercepted by the chord in the circle.

The area of the segment intercepted by the chord in the circle is $6.9{\text{in}}^{\text{2}}$ .

Given information:

The radius of the circle is $5\text{inch}$ and length of the chord is $7\text{inch}$ .

Formula used:

The formula for area of segment is $A_{\text{segment}}=A_{\text{sector}}-A_{\text{triangle}}$

Law of Cosines is $c^2=a^2+b^2-2ab\cos C$

The formula for area of sector is $A_{\text{sector}}=\frac{\theta \pi r^2}{360°}$

The formula for area of triangle is $A_{triangle}=\frac{1}{2}\times r^2\times \sin \theta$

Calculation:

Rewrite the Law of Cosine to determine the included angle

  $C={\cos }^{-1}\frac{a^2+b^2-c^2}{2ab}$

Substitute $5$ for $a$, $5$ for $b$, and $7$ for $c$ in $C={\cos }^{-1}\frac{a^2+b^2-c^2}{2ab}$ to determine the central angle.

  $\begin{array}{c}C={\cos }^{-1}\frac{5^2+5^2-7^2}{2\times 5\times 5}\\ \approx 88.85°\end{array}$

Substitute $88.85°$ for $\theta$, and $5$ for $r$ in $A_{\text{sector}}=\frac{\theta \pi r^2}{360°}$ to determine the area of sector.

  $\begin{array}{c}{A}_{\text{sector}}=\frac{\theta \pi r^2}{360°}\\ =\frac{\left(88.85\right)\times \pi \times {\left(5\right)}^2}{360°}\\ =19.4{\text{in}}^{\text{2}}\end{array}$

Substitute $5$ for $r$, and $88.85°$ for $\theta$ in $A_{triangle}=\frac{1}{2}\times r^2\times \sin \theta$ to determine the area of triangle.

  $\begin{array}{c}{A}_{triangle}=\frac{1}{2}\times r^2\times \sin \theta \\ =\frac{1}{2}\times {\left(5\right)}^2\times \sin 88.85\\ \approx 12.5{\text{in}}^{\text{2}}\end{array}$

Substitute $19.4$ for $A_{\text{sector}}$ and $12.5$ for $A_{triangle}$ in $A_{\text{segment}}=A_{\text{sector}}-A_{\text{triangle}}$ to calculate the area of segment.

  $\begin{array}{c}{A}_{\text{segment}}=A_{\text{sector}}-A_{\text{triangle}}\\ =19.4-12.5\\ =6.9{\text{in}}^{\text{2}}\end{array}$