Question

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

Answer 1

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