Answered

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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]

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