This problem represents the definition of the side lengths of an isosceles right triangle.  By definition the sides equal \(\displaystyle x\), \(\displaystyle x\), and \(\displaystyle x\sqrt{2}\).  However, if you did not remember this definition one can also find the length of the side using the Pythagorean theorem \(\displaystyle (a^{2}+b^{2}=c^{2})\).

\(\displaystyle h^{2}=x^{2}+x^{2}\)

\(\displaystyle h^{2}=2x^{2}\)

\(\displaystyle h=\sqrt{2x^{2}}\)

\(\displaystyle \rightarrow x\sqrt{2}\)

To find the length of the diagonal, given two sides of the square, we can create two equal triangles from the square. The diagonal line splits the right angles of the square in half, creating two triangles with the angles of \(\displaystyle 45\), \(\displaystyle 45\), and \(\displaystyle 90\) degrees. This type of triangle is a special right triangle, with the relationship between the side opposite the \(\displaystyle 45\) degree angles serving as x, and the side opposite the \(\displaystyle 90\) degree angle serving as \(\displaystyle x\sqrt{2}\).

Appyling this, if we plug \(\displaystyle 8\) in for \(\displaystyle x\) we get that the side opposite the right angle (aka the diagonal) is \(\displaystyle 8\sqrt{2}\)

If the area of the square is \(\displaystyle 144\), we know that each side of the square is \(\displaystyle 12\), because the area of a square is \(\displaystyle s^{2}\).

Then, the diagonal creates two \(\displaystyle 45/45/90\)special right triangles. Knowing that the sides = \(\displaystyle 12\), we can find that the hypotenuse (aka diagonal) is \(\displaystyle 12\sqrt{2}\)

In order to calculate the triangle's area, we need to find the lengths of its legs.  An isosceles triangle is a special triangle due to the values of its angles. These triangles are referred to as \(\displaystyle 45^{\circ}-45^{\circ}-90^{\circ}\) triangles and their side lengths follow a specific pattern that states that one can calculate the length of the legs of an isoceles triangle by dividing the length of the hypotenuse by the square root of 2.

\(\displaystyle x=\frac{h}{\sqrt{2}}\)

\(\displaystyle x=\frac{18\sqrt{2}in}{\sqrt{2}}\)

\(\displaystyle x=18in\)

Now we can calculate the area using the formula

 \(\displaystyle A=\frac{1}{2}bh\)

\(\displaystyle A=\frac{1}{2}*18in*18in\)

\(\displaystyle A=162in^{2}\)

Now, convert to feet.

\(\displaystyle \rightarrow \frac{162 in^{2}}{1}*\frac{1ft}{12in}*\frac{1ft}{12in}=1.125ft^{2}\)

To find the area of a triangle, multiply the base by the height, then divide by 2.  Since the short legs of an isosceles triangle are the same length, we need to know only one to know the other.  Since, a short side serves as the base of the triangle, the other short side tells us the height.

\(\displaystyle (8\cdot 8)/2=32\)

In order to calculate the triangle's perimeter, we need to find the lengths of its legs.  An isosceles triangle is a special triangle due to the values of its angles. These triangles are referred to as \(\displaystyle 45^{\circ}-45^{\circ}-90^{\circ}\) triangles and their side lengths follow a specific pattern that states that one can calculate the length of the legs of an isoceles triangle by dividing the length of the hypotenuse by the square root of 2.

\(\displaystyle x=\frac{h}{\sqrt{2}}\)

\(\displaystyle x=\frac{8}{\sqrt{2}}\)

\(\displaystyle x=4\sqrt{2}\)

Now we can calculate the perimeter by doubling \(\displaystyle x\) and adding \(\displaystyle h\).

\(\displaystyle P=2x+h\)

\(\displaystyle P=(2*4\sqrt{2})+8cm\)

\(\displaystyle P=19.31cm\)

An isosceles triangle is a special triangle due to the values of its angles.  These triangles are referred to as \(\displaystyle 45^{\circ}-45^{\circ}-90^{\circ}\) triangles and their side lenghts follow a specific pattern that states you can calculate the length of the hypotenuse of an isoceles triangle by multiplying the length of one of the legs by the square root of 2.

\(\displaystyle h=x\sqrt{2}\)

\(\displaystyle h=14cm\sqrt{2}\)

\(\displaystyle h=19.80cm\)

Now we can calculate the perimeter by doubling \(\displaystyle x\) and adding \(\displaystyle h\).

\(\displaystyle P=2x+h\)

\(\displaystyle P=(14cm*2)+19.80cm\)

\(\displaystyle P=47.8cm\)

When a triangle has two angles equal to \(\displaystyle 45^\circ\), it must be a \(\displaystyle 45^\circ:45^\circ:90^\circ\) isosceles right triangle.

The pattern for the sides of a \(\displaystyle 45^\circ:45^\circ:90^\circ\) is \(\displaystyle x:x:x\sqrt{2}\).

Since two sides are equal to \(\displaystyle 8\), this triangle will have sides of \(\displaystyle 8:8:8\sqrt{2}\).

Add them all together to get \(\displaystyle 16+8\sqrt{2}\).

An isosceles triangle is basically two right triangles stuck together. The isosceles triangle has a base of 6, which means that from the midpoint of the base to one of the angles, the length is 3. Now, you have a right triangle with a base of 3 and a height of 4. The hypotenuse of this right triangle, which is one of the two congruent sides of the isosceles triangle, is 5 units long (according to the Pythagorean Theorem).

The total perimeter will be the length of the base (6) plus the length of the hypotenuse of each right triangle (5).

5 + 5 + 6 = 16