What is the perimeter of a triangle with corners at (7 ,6 )(7,6), (4 ,5 )(4,5), and (3 ,1 )(3,1)?

1 Answer
Aug 10, 2017

Perimeter=sqrt10+sqrt17+sqrt41Perimeter=10+17+41

Explanation:

Equation for finding distance between 22 coordinates == sqrt((x_2-x_1)^2+(y_2-y_1)^2)(x2x1)2+(y2y1)2

Let,
(7,6)=(7,6)=point AA

(4,5)=(4,5)=point BB

(3,1)=(3,1)=point CC

Distance between ABAB

x_1=7,x_2=4,y_1=6,y_2=5x1=7,x2=4,y1=6,y2=5

AB=sqrt((4-7)^2+(5-6)^2)AB=(47)2+(56)2

=>sqrt((-3)^2+(-1)^2(3)2+(1)2

=>sqrt(9+1)9+1

=>sqrt1010

AB=sqrt10AB=10

Distance between BCBC

x_1=4,x_2=3,y_1=5,y_2=1x1=4,x2=3,y1=5,y2=1

BC=sqrt((3-4)^2+(1-5)^2)BC=(34)2+(15)2

=>sqrt((-1)^2+(-4)^2(1)2+(4)2

=>sqrt(1+16)1+16

sqrt1717

BC=sqrt17BC=17

Distance between ACAC

x_1=7,x_2=3,y_1=6,y_2=1x1=7,x2=3,y1=6,y2=1

AC=sqrt((3-7)^2+(1-6)^2)AC=(37)2+(16)2

=>sqrt((-4)^2+(-5)^2(4)2+(5)2

=>sqrt(16+25)16+25

=>sqrt4141

AC=sqrt41AC=41

Perimeter of a triangle is the sum of its 33 sides, that AB+BC+AC.AB+BC+AC.

Perimeter=sqrt10+sqrt17+sqrt41Perimeter=10+17+41