Question

find the perimeter of triangle with vertices ( 0,8 ) ( 6,0 ) and origin ( 9,3 ) ( 1, - 3 ) and orgin

Answer 1

Use the distance formula
First triangle:
(0,0) (0,8) (6,0)
the 3 sides are
$\sqrt{ (0-0)^{2} + (0-8)^{2} } = 8$
$\sqrt{ (0-0)^{2} + (0-6)^{2} } = 6$
$\sqrt{ (0-6)^{2} + (8-0)^{2} } = 10$
Perimeter = 24

Second Triangle:
(0,0) (9,3) (1,-3)
$\sqrt{ (0-9)^{2} + (0-3)^{2} } =3\ \sqrt{10}$
$\sqrt{ (0-1)^{2} + (0-(-3))^{2} } = \sqrt{10}$
$\sqrt{ (9-1)^{2} + (3-(-3))^{2} } = 10$
Perimeter = $10 + 4 \sqrt{10}$

Answer 2

Answer:

oiiiihyuyujfr867rk,ud5,t,i5k7oy.r6lo$\geq \leq x^{2} x^{2} \sqrt[n]{x}$→³∴∧∨

Step-by-step explanation: