Question

Triangle NOP, with vertices N(3,5), (7,9), and P(2,8), is drawn on the coordinategrid below.109N43TOWhat is the area, in square units, of triangle NOP?

Answer 1

To find the area we will use the heron's formula

`\text{Area}=\sqrt[]{p(p-a)(p-b)(p-c)}`

where p is half of the perimeter

a,b,c are the lengthof the sides of the triangle

First, we need to to find the lengths using the distance formula

`|d|=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}`

Let N0 be side a, OP be side b and PN be side c

Let's find N0 (side a)

N(3,5) 0(7,9)

x₁=3 y₁=5 x₂=7 y₂=9

substitute the values into the formula and evaluate

`|NO|=\sqrt[]{(7-3)^2+(9-5)^2}`

`=\sqrt[]{4^2+4^2}`
`=\sqrt[]{16+16}=\sqrt[]{32}\approx5.6569`

Next, is to find OP (side b)

0(7, 9) P(2,8)

x₁=7 y₁=9 x₂=2 y₂=8

substitute the values into the formula and evaluate

`|0P|=\sqrt[]{(2-7)^2+(8-9)^2}`

`=\sqrt[]{(-5)^2+(-1)^2}`
`=\sqrt[]{25+1}=\sqrt[]{26}\approx5.099`

Next is to find PN (side c)

P(2,8) N(3,5)

x₁=2 y₁=8 x₂=3 y₂=5

`|PN|=\sqrt[]{(3-2)^2+(5-8)^2}`

`=\sqrt[]{(1)^2+(-3)^2}`

`=\sqrt[]{1+9}=\sqrt[]{10}\approx3.1623`

We can go ahead and find p

`p=((5.6569+5.099+3.1623))/(2)`
`\approx6.9591`

p=6.9591 a=5.6569 b=5.099 c=3.1623

substitute the values to get your area

`A=\sqrt[]{6.9591(6.9591-5.6569)(6.9591-5.099)(6.9591-3.1623)}`
`=\sqrt[]{6.9591(1.3022)(1.8601)(3.7968)}`

`=\sqrt[]{64}`
`=8\text{ square units}`