Question

In isosceles ΔABC, AC = BC, AB = 6 in, CD ⊥ AB, and CD = 3 in. Find the perimeter of the isosceles triangle.

Answer 1

Answer:

$4\sqrt{3}+6$

Step-by-step explanation:

We know that the base is 6 and the altitude is $\sqrt{3}$

We need to know the sides of the triangle

The triangle is already divided into two by the altitude, the new base will be 3 because 6 divided by 2 is 3, and the side of the smaller triangle will be $\sqrt{3}$

So we use the pythagoream thereom: $(\sqrt{3})^2+3^2=c^2$

Now since we are squaring a square root, it goes back to 3, and we know $3^2$ equals 9: $3+9= c^2$

Now we solve: $12= c^2$

                        $\sqrt{12}=c$

Now we know the side of the triangle ABC

Since it is an isosceles, both sides are  $\sqrt{12}$

Now we add all sides to get our perimeter: P= $\sqrt{12}+\sqrt{12}+6$

We know that $\sqrt{12}$ + $\sqrt{12}$ = $\sqrt{24}$ = $4\sqrt{3}$ .......when we factor

Which is: $4\sqrt{3}+6$

Answer 2

Answer:

Step-by-step explanation:

Consider the given triangle ABC, we have AC = BC, AB = 6 in, CD ⊥ AB, and CD = 3 in. Using ΔCDB, we have

$(CB)^{2}=(CD)^{2}+(DB)^{2}$

$(CB)^{2}=(3)^{2}+(3)^{2}$

$CB=\sqrt{9+9}$

$CB=\sqrt{18}in$

Therefore, $CB=CA=\sqrt{18}in$ (Given)

Now, Perimeter of an isosceles triangle is given By: 2a+b

⇒$2(\sqrt{18})+6=2(4.24)+6=8.48+6=14.48in$

which is the required perimeter of isosceles triangle.