Find the perimeter of the triangle with vertices (0,,0) and origin ; (9,,-3) and origin .

12

Step-by-step explanation:

=√(x2-x1)^2+(y2-y1)^2

=√16+√16+9+√9

=12

12cm is the answer

Step-by-step explanation:

24

Step-by-step explanation:

Let the triangle be ABC where

  A(0,0) ;  B(6,0) ;  C(0,8)

AB=

BC=

CA=

Perimeter of triangle= AB + BC + CA = 6 + 10 + 8 = 24

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Hello mate,

If we draw the triangle with vertices (0,1),(0,0),(1,0),

Then, we get that the height of the triangle is 1 units and the base is also 1 unit,

so,The area of the triangle =1/2*base*height

=1/2*1 unit*1 unit=0.5 unit square,

Thanks for questioning

and,Please mark me as brainliest..

Hey mate

Here goes your answer

Let A(0,4),B(0,0)and C(3,0)

By distance formula

AB = √(0-4)^2+(0-0)^2

AB=√16

AB=4

BC=√(0-3)^2+(0-0)^2

BC=√9=3

AC=√(0-4)^2+(0-3)^2

AC=√16+9

AC=√25=5

By heron's formula

=√6(6-5)(6-4)(6-3)

=√6*1*2*3

=√6*6

=6 unit sq is your answer.

Hope it helps you

Step-by-step explanation:

let A(0,0) B(6,0) C(0,8)

AB = √{(0-6)^2 + (0-0)^2}

= √ (-6)^2

= 6 units

BC = √{(6-0)^2 + (0-8)^2}

= √{(6)^2 + -8)^2}

= √36 + 64

= √100

= 10 units

AC = √{(0-0)^2 + (0-8)^2}

= √(-8)^2

= 8 units

perimeter = AB + BC + AC

= 6 + 10 + 8

= 24 units

Step-by-step explanation:

This is a triangle which has sides of 6cm and 8cm and is a right angled triangle.

By using Pythagoras theorem,

Let the side to be found out be x.

Therefore,

Therefore Perimeter :

8+6+10=24cm

answer:

don't post useless questions

answer:

i can't get the question