Perimeter of a Triangle: The Perimeter of a Triangle

In order to find the perimeter of a triangle, we first need to find all of its sides.

Two sides have already been given leaving only one remaining side to find.

We can use the Pythagorean Theorem.
$AB^2+BC^2=AC^2$
We insert all of the known data:

$AC^2=7^2+3^2$
$AC^2=49+9=58$
We extract the square root:

$AC=\sqrt{58}$
Now that we have all of the sides, we can add them up and thus find the perimeter:
$\sqrt{58}+7+3=\sqrt{58}+10$

This problem involves determining the lengths of the legs of a triangle whose perimeter is 12 cm, given that one side is 5 cm. To solve, consider the apparent context that implies a right triangle.

First, let's denote the three sides of the triangle as $a$, $b$, and $c$, where $c = 5$ cm.

Considering the perimeter formula:

$a + b + c = 12$

Since $c$ is 5 cm, the equation becomes:

$a + b + 5 = 12$

Solving for $a + b$:

$a + b = 7$

Assuming it is a right triangle with the side length of 5 cm as the hypotenuse:

$c^2 = a^2 + b^2$

Where $c = 5$, the equation is:

$5^2 = a^2 + b^2$
$25 = a^2 + b^2$

We need the integers $a$ and $b$ that satisfy both $a + b = 7$ and $a^2 + b^2 = 25$.

To trial integer pairs from $a + b = 7$:

- If $a = 3$, then $b = 4$.

Check $a = 3$ and $b = 4$ in the Pythagorean condition:

$3^2 + 4^2 = 9 + 16 = 25$

Hence, the pair satisfies both conditions.

Therefore, the lengths of the legs are $3 \, \text{cm}$ and $4 \, \text{cm}$.

We calculate the perimeter of the triangle:

$12+4\sqrt{5}=4+AC+BC$

As we want to find the hypotenuse (BC), we isolate it:

$12+4\sqrt{5}-4-AC=BC$

$BC=8+4\sqrt{5}-AC$

Then calculate AC using the Pythagorean theorem:

$AB^2+AC^2=BC^2$

$4^2+AC^2=(8+4\sqrt{5}-AC)^2$

$16+AC^2=(8+4\sqrt{5})^2-2\times AC(8+4\sqrt{5})+AC^2$

We then simplify the two:$AC^2$

$16=8^2+2\times8\times4\sqrt{5}+(4\sqrt{5})^2-2\times8\times AC-2AC4\sqrt{5}$

$16=64+64\sqrt{5}+16\times5-16AC-8\sqrt{5}AC$

$16AC+8\sqrt{5}AC=64+64\sqrt{5}+16\times5-16$

$AC(16+8\sqrt{5})=128+64\sqrt{5}$

$AC=\frac{128+64\sqrt{5}}{16+8\sqrt{5}}=\frac{8(16+8\sqrt{5})}{16+8\sqrt{5}}$

We simplify to obtain:

$AC=8$

Now we can replace AC with the value we found for BC:

$BC=8+4\sqrt{5}-AC$

$BC=8+4\sqrt{5}-8=4\sqrt{5}$