Question

Triangle ABC is an isosceles triangle in which AB = AC. What is the perimeter of △ABC?

Answer 1

Answer:

Option C.

Step-by-step explanation:

Assume that the below figure attached with this question.

Distance formula:

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

From the below figure it is clear that the vertices of triangle are A(-1,6), B(-1,1) and C(2,2).

Using distance formula we get

$AB=\sqrt{\left(-1-\left(-1\right)\right)^2+\left(1-6\right)^2}\Rightarrow \sqrt{0+25}=5$

Similarly,

$BC=\sqrt{\left(2-\left(-1\right)\right)^2+\left(2-1\right)^2}=\sqrt{10}$

$AC=\sqrt{\left(2-\left(-1\right)\right)^2+\left(2-6\right)^2}=5$

The perimeter of △ABC is

$perimeter=AB+BC+AC$

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

$perimeter=10+\sqrt{10}$

The perimeter of △ABC is $10+\sqrt{10}$.

Therefore, the correct option is C.

Answer 2

Answer:

It is definitely B

Step-by-step explanation:

$10 + \sqrt{10} units$