Answered

Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

what is the area of a figure with vertices(1,1), (8,1) , and (5,5)?

Sagot :

luana
It will be a triangle
Calculating sides of this triangle
A(1,1)
B(8,1)
C(5,5)
side a-AB=[tex] \sqrt{(8-1)^{2}+( 1-1)^{2} ) }= \sqrt{ 7^{2} }=7 [/tex]
side b-AC=[tex] \sqrt{(5-1)^{2}+( 5-1)^{2} ) }= \sqrt{ 4^{2}+4^{2} }=[tex] \sqrt{36} [/tex]=6 [/tex]
side c-BC=[tex] \sqrt{(5-8)^{2}+( 5-1)^{2} ) }= \sqrt{ (-3)^{2}+4^{2} }=[tex] \sqrt{25} [/tex]=5 [/tex]
Area of the triangle from Herone formula:
A=[tex] \sqrt{p(p-a)(p-b)(p-c)} [/tex]
p=[tex] \frac{1}{2}(a+b+c) [/tex]=[tex] \frac{1}{2}(7+6+5) [/tex]=9
A=[tex] \sqrt{9(9-7)(9-6)(9-5)} [/tex]=[tex] \sqrt{9*2*3*4} [/tex]=[tex] \sqrt{216} [/tex]
Area is [tex] \sqrt{216} [/tex]

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.