Question

On a coordinate plane, triangle A B C is shown. Point A is at (3, 4), point B is at (negative 5, negative 2), and point C is at (5, negative 2). In the diagram, AB = 10 and AC = 2 StartRoot 10 EndRoot. What is the perimeter of △ABC? 10 units 10 + 2 StartRoot 10 EndRoot units 20 units 20 + 2 StartRoot 10 EndRoot units

Answer 1

Answer:

its d on e2020

Step-by-step explanation:

Answer 2

Answer:

The perimeter of Δ ABC is 20 + 2$\sqrt{10}$ units ⇒ Last answer

Step-by-step explanation:

The perimeter of any triangle is the sum of the lengths of its three sides

The formula of distance between two points is $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

In Δ ABC

∵ A = (3 , 4) , B = (-5 , -2) , C = (5 , -2)

∵ AB = 10 units

∵ AC = 2$\sqrt{10}$

- To find its perimeter find the length of BC

∵ $x_{1}$ = -5 and $y_{1}$ = -2

∵ $x_{2}$ = 5 and $y_{2}$ = -2

- By using the formula above

∴ $BC=\sqrt{(5--5)^{2}+(-2--2)^{2}}=\sqrt{(5+5)^{2}+(-2+2)^{2}}$

∴ $BC=\sqrt{(10)^{2}+(0)^{2}}=\sqrt{100}$

∴ BC = 10 units

To find the perimeter add the lengths of the three sides

∵ P = AB + BC + AC

∴ P = 10 + 10 + 2$\sqrt{10}$

- Add like terms

∴ P = 20 + 2$\sqrt{10}$

The perimeter of Δ ABC is 20 + 2$\sqrt{10}$ units