+0  
 
0
80
1
avatar+214 

Help plssss

chitsuii  Jun 28, 2018

Best Answer 

 #1
avatar+2190 
+1

One way to approach this problem is to find the area of the circle and subtract the area of the minor sector and the area of the three congruent triangles. 

 

Let's find the area of the circle:

 

\(A_{\circ}=\pi*6^2\) Use the formula for the area of a circle.
\(A_{\circ}=36\pi\text{ft}^2\)  
   

 

Let's find the area of the minor sector:

 

\(A_{\text{sector}}=\frac{1}{4}A_{\circ}\) This minor sector is really just a quarter of the area of the entire circle.
\(A_{\text{sector}}=\frac{1}{4}*36\pi\text{ft}^2\)  
\(A_{\text{sector}}=9\pi\text{ft}^2\)  
   

 

Let's find the area of the three congruent triangles that are right and isosceles:

 

\(A_{\triangle}=\frac{1}{2}*6*6\) It is not too difficult to find the area of these triangles since they are all right. 
\(A_{\triangle}=18\text{ft}^2\)  
\(3A_{\triangle}=3*18\text{ft}^2\) Thus far, I have calculated the area of one triangle, but we need to calculate the area of all three. 
\(3A_{\triangle}=54\text{ft}^2\)  

 

Now that we know everything, let's find the area of the three minor segments of this particular circle:

 

\(A_{\text{segments}}=A_{\circ}-A_{\text{sector}}-3A_{\triangle}\) Do the substitution. 
\(A_{\text{segments}}=(36\pi-9\pi-54)\text{ft}^2\) Combine like terms.
\(A_{\text{segments}}=(27\pi-54)\text{ft}^2\) Match this with the answer choices above.
   
TheXSquaredFactor  Jun 28, 2018
 #1
avatar+2190 
+1
Best Answer

One way to approach this problem is to find the area of the circle and subtract the area of the minor sector and the area of the three congruent triangles. 

 

Let's find the area of the circle:

 

\(A_{\circ}=\pi*6^2\) Use the formula for the area of a circle.
\(A_{\circ}=36\pi\text{ft}^2\)  
   

 

Let's find the area of the minor sector:

 

\(A_{\text{sector}}=\frac{1}{4}A_{\circ}\) This minor sector is really just a quarter of the area of the entire circle.
\(A_{\text{sector}}=\frac{1}{4}*36\pi\text{ft}^2\)  
\(A_{\text{sector}}=9\pi\text{ft}^2\)  
   

 

Let's find the area of the three congruent triangles that are right and isosceles:

 

\(A_{\triangle}=\frac{1}{2}*6*6\) It is not too difficult to find the area of these triangles since they are all right. 
\(A_{\triangle}=18\text{ft}^2\)  
\(3A_{\triangle}=3*18\text{ft}^2\) Thus far, I have calculated the area of one triangle, but we need to calculate the area of all three. 
\(3A_{\triangle}=54\text{ft}^2\)  

 

Now that we know everything, let's find the area of the three minor segments of this particular circle:

 

\(A_{\text{segments}}=A_{\circ}-A_{\text{sector}}-3A_{\triangle}\) Do the substitution. 
\(A_{\text{segments}}=(36\pi-9\pi-54)\text{ft}^2\) Combine like terms.
\(A_{\text{segments}}=(27\pi-54)\text{ft}^2\) Match this with the answer choices above.
   
TheXSquaredFactor  Jun 28, 2018

16 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.