Question

find the perimeter of triangle with vertices of (0,5) (0,0) (12,2)???​

Answer 1

Answer:

29.5348419

Step-by-step explanation:

-We know that,

distance formula = √(y2-y1)²+(x2-x1)²

- so,

AC=√(2-5)²+(12-0)² =√(-3)²+(12)² =√9+144

= √153

AB=√(0-5)²+((0-0)² =√(-5)²+(0)² =√25

= 5

BC=√(2-0)²+(12-0)² =√(2)²+(12)² =√4+144

= √148

The parameter of triangle

= AC+AB+BC

=12.3693169 + 5 + 12.1655251

=29.5348419

~30

Answer 2

Let ABC be a triangle having A(0,4);B(0,0);C(3,0)

Using distance formula, we get AB=

$=(0−0)2+(0−4)2=16=4$

$BC=(3−0)2+(0−0)2=9=3$

CA=

$=(0−3)2+(4−0)2=25=5$

Perimeter of △ABC=AB+BC+CA=4+3+5=12

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