An isoceles right triangle is another way of saying that the triangle is a \(\displaystyle 45-45-90\) triangle. 

Now, recall the Pythagorean theorem:

\(\displaystyle \text{base}^2+\text{height}^2=\text{hypotenuse}^2\)

Because we are working with a \(\displaystyle 45-45-90\) triangle, the base and the height have the same length. We can rewrite the above equation as the following:

\(\displaystyle \text{height}^2+\text{height}^2=\text{hypotenuse}^2\)

\(\displaystyle 2(\text{height})^2=\text{hypotenuse}^2\)

\(\displaystyle \text{height}^2=\frac{\text{hypotenuse}^2}{2}\)

\(\displaystyle \text{height}=\sqrt{\frac{\text{hypotenuse}^2}{2}}\)

Simplify.

\(\displaystyle \text{height}=\frac{\text{hypotenuse}}{\sqrt2}\)

Multiply the fraction by one in the form of:

\(\displaystyle 1\rightarrow \frac{\sqrt{2}}{\sqrt{2}}\)

Substitute.

\(\displaystyle \text{height}=\frac{\text{hypotenuse}}{\sqrt2}\times \frac{\sqrt{2}}{\sqrt{2}}\)

Solve.

\(\displaystyle \text{height}=\frac{\text{hypotenuse}(\sqrt2)}{2}{}\)

Now, substitute in the value of the hypotenuse to find the height for the given triangle.

\(\displaystyle \text{height}=\frac{144\sqrt2}{2}\)

Simplify.

\(\displaystyle \text{height}=72\sqrt2\)

An isoceles right triangle is another way of saying that the triangle is a \(\displaystyle 45-45-90\) triangle. 

Now, recall the Pythagorean theorem:

\(\displaystyle \text{base}^2+\text{height}^2=\text{hypotenuse}^2\)

Because we are working with a \(\displaystyle 45-45-90\) triangle, the base and the height have the same length. We can rewrite the above equation as the following:

\(\displaystyle \text{height}^2+\text{height}^2=\text{hypotenuse}^2\)

\(\displaystyle 2(\text{height})^2=\text{hypotenuse}^2\)

\(\displaystyle \text{height}^2=\frac{\text{hypotenuse}^2}{2}\)

\(\displaystyle \text{height}=\sqrt{\frac{\text{hypotenuse}^2}{2}}\)

Simplify.

\(\displaystyle \text{height}=\frac{\text{hypotenuse}}{\sqrt2}\)

Multiply the fraction by one in the form of:

\(\displaystyle 1\rightarrow \frac{\sqrt{2}}{\sqrt{2}}\)

Substitute.

\(\displaystyle \text{height}=\frac{\text{hypotenuse}}{\sqrt2}\times \frac{\sqrt{2}}{\sqrt{2}}\)

Solve.

\(\displaystyle \text{height}=\frac{\text{hypotenuse}(\sqrt2)}{2}{}\)

Now, substitute in the value of the hypotenuse to find the height for the given triangle.

\(\displaystyle \text{height}=\frac{95\sqrt2}{2}\)

An isoceles right triangle is another way of saying that the triangle is a \(\displaystyle 45-45-90\) triangle. 

Now, recall the Pythagorean theorem:

\(\displaystyle \text{base}^2+\text{height}^2=\text{hypotenuse}^2\)

Because we are working with a \(\displaystyle 45-45-90\) triangle, the base and the height have the same length. We can rewrite the above equation as the following:

\(\displaystyle \text{height}^2+\text{height}^2=\text{hypotenuse}^2\)

\(\displaystyle 2(\text{height})^2=\text{hypotenuse}^2\)

\(\displaystyle \text{height}^2=\frac{\text{hypotenuse}^2}{2}\)

\(\displaystyle \text{height}=\sqrt{\frac{\text{hypotenuse}^2}{2}}\)

Simplify.

\(\displaystyle \text{height}=\frac{\text{hypotenuse}}{\sqrt2}\)

Multiply the fraction by one in the form of:

\(\displaystyle 1\rightarrow \frac{\sqrt{2}}{\sqrt{2}}\)

Substitute.

\(\displaystyle \text{height}=\frac{\text{hypotenuse}}{\sqrt2}\times \frac{\sqrt{2}}{\sqrt{2}}\)

Solve.

\(\displaystyle \text{height}=\frac{\text{hypotenuse}(\sqrt2)}{2}{}\)

Now, substitute in the value of the hypotenuse to find the height for the given triangle.

\(\displaystyle \text{height}=\frac{52\sqrt2}{2}\)

Simplify.

\(\displaystyle \text{height}=26\sqrt2\)

To find the height, use the Pythagorean Theorem. One of the legs is the missing side, and the other is 1.5, half of 3. The hypotenuse is 5:

\(\displaystyle (1.5)^2 + x^2 = 5^ 2\)

\(\displaystyle 2.25 + x^2 = 25\)

\(\displaystyle x^2 = 22.75\)

\(\displaystyle x \approx 4.77\)

\(\displaystyle Area\;\Delta=\frac{1}{2}(base)(height)\)

\(\displaystyle \frac{1}{2}(3)(3)=\frac{9}{2}=4.5\)

The formula for the area of a trianlge is A = base * height * (1/2).

We're lucky here, because the question gives us all of the values we need.  We simply need to plug them in:

A = base * height * (1/2) = 12 * 24 * (1/2) = 12 * 12 = 144

The formula for the area of a triangle, right or not, is one half the base times height.

In this case, they are both \(\displaystyle 4 \text{ in.}\) Therefore, the respective values are entered, yielding:

 \(\displaystyle \frac{1}{2}(4\text{ in.})(4\text{ in.})=8 \text{ in.}^2\)

The area of a triangle is:

\(\displaystyle A=\frac{b*h}{2}\) where b=base, h=height, and A=area

\(\displaystyle A=\frac{20\sqrt{2}*20\sqrt{2}}{2}=\frac{400\sqrt{4}}{2}=200\sqrt{4}=200*2=400\)

The key to finding the area of our triangle is to reaize that it is isosceles and therefore is a 45-45-90 triangle; therefore, we know the legs of our triangle are congruent and that each can be found by dividing the length of the hypotenuse by \(\displaystyle \sqrt{2}\).

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

Rationalizing the denominator simplifies our result; however, we are interested in the area, not just the length of a leg; we remember that the formula for the area of a triangle is

\(\displaystyle A = \frac{1}{2}bh\)

where \(\displaystyle b\) is the base and \(\displaystyle h\) is the height; however, in our right triangle, the base and height are simply the two legs; therefore, we can calculate the area by substituting.

\(\displaystyle A=\frac{1}{2}(\frac{8\sqrt2}{2})(\frac{8\sqrt2}{2})=\frac{128}{8}=16\)

To find the area of a triangle, we must use \(\displaystyle Area=\frac{1}{2}b\cdot h\) where b=base and h=height. 

In the problem, the only information given is what type the triangle is and what its hypotenuse is. 

Given the area equation, the problem hasn't given any numbers that can be substituted into the equation to solve for an area. This means that the hypotenuse value must be used to determine the height and the base. 

Because this is a 45/45/90 triangle, this means that it is also isosceles. Therefore, we can logic out that the base and the height must be the same. 

The missing sides can be calulated in one of two ways:

1. Using the Pythagorean Theorem \(\displaystyle a^2+b^2=c^2\)

2. Or using 

If we were to use the Pythagorean Theorem, since we've already determined that b=h, that mean a=b in the equation. Let's say that \(\displaystyle a=b=s\). 

That means the Pythagorean Theorem can be rewritten as:

\(\displaystyle s^2+s^2=c^2\)

\(\displaystyle {\color{Blue} 2s^2=c^2}\)

Now to substitute in the value of c to solve for the height and base. 

\(\displaystyle 2s^2=(5\sqrt{3})^2\)

\(\displaystyle 2s^2=25\sqrt{9}\)

\(\displaystyle 2s^2=75\)

\(\displaystyle s^2=\frac{75}{2}\)

\(\displaystyle \sqrt{s^2}=\sqrt{\frac{75}{2}}\)

\(\displaystyle s=\frac{5\sqrt{6}}{2}\)

Now that we have the base and the height, we can substitute the values into the area equation and get the triangle's area. 

\(\displaystyle Area=\frac{1}{2}\cdot\frac{5\sqrt{6}}{2}\cdot\frac{5\sqrt{6}}{2}\)

\(\displaystyle Area=\frac{1}{2}\cdot\frac{(25\cdot 6)}{4}\)

\(\displaystyle Area = \frac{1}{2}\cdot\frac{150}{4}\)

\(\displaystyle Area = \frac{1}{2}\cdot \frac{75}{2} = {\color{Blue} \frac{75}{4}}\)