Expand your knowledge base with the help of IDNLearn.com's extensive answer archive. Ask anything and receive comprehensive, well-informed responses from our dedicated team of experts.

What is the area of FGH? (Please help it’s due tmrw !! I’ll mark you brainliest)

What Is The Area Of FGH Please Help Its Due Tmrw Ill Mark You Brainliest class=

Sagot :

Answer:

102.0625 square units

Step-by-step explanation:

First determine the length of each side using the distance formula  for two points. Then use Heron's formula to determine the area of a triangle given three sides

Distance Formula

The distance between two points is the length of the path connecting them

The distance between points (x₁, y₁) and (x₂, y₂) is given by the Pythagorean theorem:

[tex]d = \sqrt {(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}[/tex]

Let's compute the lengths of the sides FH, FG and HG

The three vertices are F(-2, 5)   G(7, -10) and H(-9, -6)  as indicated on the graph

So length FG between (-2,5) and (7,-10)

[tex]FG= \sqrt {(7 - (-2))^2 + (-10 - 5)^2}[/tex]

[tex]= \sqrt {(9)^2 + (-15)^2}[/tex]

[tex]= \sqrt {{81} + {225}}[/tex]

[tex]= \sqrt {306}[/tex]

FG [tex]= 17.492856[/tex]  (round to 17.5)

Length FH between(-2,5) and (-9, -6) is

[tex]FH = \sqrt {(-9 - (-2))^2 + (-6 - 5)^2}[/tex]

[tex]= \sqrt {(-7)^2 + (-11)^2}[/tex]

[tex]= \sqrt {{49} + {121}}[/tex]

[tex]= \sqrt {170}[/tex]

FH [tex]= 13.038405[/tex]  (can be rounded to 13.04

Length GH between (7, -10) and (-9, -6) is

[tex]GH = \sqrt {(-9 - 7)^2 + (-6 - (-10))^2}[/tex]

[tex]= \sqrt {(-16)^2 + (4)^2}[/tex]

[tex]= \sqrt {{256} + {16}}[/tex]

[tex]= \sqrt {272}[/tex]

GH [tex]= 16.492423[/tex]   (can be rounded to 16.5)



Determining the area of a triangle given 3 sides

Heron's formula allows us to find the area of a triangle given 3 sides If the sides are a, b and c the general form of Heron's formula is

[tex]Area = \sqrt {s(s-a)(s-b)(s-c)}[/tex]

where s is the semi-perimeter = [tex]\frac{a+b+c}{2}[/tex]

Substituting values we get

s = [tex]\frac{17.5+13.04+16.5}{2} = 23.52[/tex]

[tex]Area = \sqrt {23.52(23.52-17.5)(23.52-13.04)(23.52-16.5)}[/tex]

[tex]= \sqrt{23.52\cdot6.02\cdot10.48\cdot7.02}[/tex]

[tex]102.0625[/tex]  square units