Question

Triangle ABC is an isosceles triangle in which side

Answer 1

Answer:

10 + √10 units

Step-by-step explanation:

Answer 2

Answer:

Option B.

Step-by-step explanation:

Consider the below figure attached with this question.

It is given that triangle ABC is an isosceles triangle.

From the below figure it is clear that the vertices of triangle are A(-2,-4), B(2,-1) and C(3,-4).

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using this formula, we get

$AB=\sqrt{(2-(-2))^2+(-1-(-4))^2}=\sqrt{16+9}=5$

$BC=\sqrt{(3-2)^2+(-4-(-1))^2}=\sqrt{1+9}=\sqrt{10}$

$AC=\sqrt{(3-(-2))^2+(-4-(-4))^2}=\sqrt{25}=5$

Now,

Perimeter of triangle ABC = AB + BC + AC

                        $=5+\sqrt{10}+5$

                        $=10+\sqrt{10}$

So, perimeter of triangle ABC is $10+\sqrt{10}$ units.

Therefore, the correct option is B.