Question

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

Answer 1

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=√(9+9)`

`CB=√(18)in`

Therefore,
`CB=CA=√(18)in` (Given)

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

⇒
`2(√(18))+6=2(4.24)+6=8.48+6=14.48in`

which is the required perimeter of isosceles triangle.

Answer 2

`4√(3)+6`

Explanation:

We know that the base is 6 and the altitude is
`√(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
`√(3)`

So we use the pythagoream thereom:
`(√(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`

`√(12)=c`

Now we know the side of the triangle ABC

Since it is an isosceles, both sides are
`√(12)`

Now we add all sides to get our perimeter: P=
`√(12)+√(12)+6`

We know that
`√(12)` +
`√(12)` =
`√(24)` =
`4√(3)` .......when we factor

Which is:
`4√(3)+6`