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

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boffeemadrid

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

[tex](CB)^{2}=(CD)^{2}+(DB)^{2}[/tex]

[tex](CB)^{2}=(3)^{2}+(3)^{2}[/tex]

[tex]CB=\sqrt{9+9}[/tex]

[tex]CB=\sqrt{18}in[/tex]

Therefore, [tex]CB=CA=\sqrt{18}in[/tex] (Given)

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

⇒[tex]2(\sqrt{18})+6=2(4.24)+6=8.48+6=14.48in[/tex]

which is the required perimeter of isosceles triangle.

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IThinkIKnowTheAnswer

Answer:

[tex]4\sqrt{3}+6[/tex]

Step-by-step explanation:

We know that the base is 6 and the altitude is [tex]\sqrt{3}[/tex]

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 [tex]\sqrt{3}[/tex]

So we use the pythagoream thereom: [tex](\sqrt{3})^2+3^2=c^2[/tex]

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

Now we solve: [tex]12= c^2[/tex]

                        [tex]\sqrt{12}=c[/tex]

Now we know the side of the triangle ABC

Since it is an isosceles, both sides are  [tex]\sqrt{12}[/tex]

Now we add all sides to get our perimeter: P= [tex]\sqrt{12}+\sqrt{12}+6[/tex]

We know that [tex]\sqrt{12}[/tex] + [tex]\sqrt{12}[/tex] = [tex]\sqrt{24}[/tex] = [tex]4\sqrt{3}[/tex] .......when we factor

Which is: [tex]4\sqrt{3}+6[/tex]

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