Use the Pythagorean Theorem to find the other leg.

$\displaystyle {}a^2+b^2=c^2$

$\displaystyle a^2+3^2=4^2$

$\displaystyle a^2=7$

$\displaystyle a=\sqrt7$

The length of the given leg is 3, and the unknown leg is $\displaystyle \sqrt7$.

Use the area of a triangle formula and solve.

$\displaystyle A=\frac{bh}{2}= \frac{3\sqrt7}{2}$

Write the formula for the Pythagorean theorem.

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

In an isosceles right triangle, both legs of the right triangle are equal.  

$\displaystyle a=b$

Substitute the either variable and the known hypotheuse and determine the side length.

$\displaystyle b^2+b^2 =1$

$\displaystyle 2b^2=1$

$\displaystyle b^2=\frac{1}{2}$

$\displaystyle b=\sqrt\frac{1}{2}$

This length represents both the base and the height of the triangle. Write the area of a triangle and substitute to solve for the area.

$\displaystyle A=\frac{bh}{2}=\frac{\sqrt\frac{1}{2}\sqrt\frac{1}{2}}{2}=\frac{\frac{1}{2}}{2}= \frac{1}{4}$

Given that:

C=90°

B=45°

a=5

c=$\displaystyle \sqrt{50}$

$\displaystyle a^{2}+b^{2}=c^{2}$

Therefore...

$\displaystyle 5^{2}+b^{2}=(\sqrt{50})^{2}$

$\displaystyle 25+b^{2}=50$

$\displaystyle b^{2}=50-25$

$\displaystyle b^{2}=25$

$\displaystyle \sqrt{b^{2}} =\sqrt{25}$

$\displaystyle b=5$

All angles of a triangle add up to 180°.

$\displaystyle A+45^{\circ}+90^{\circ}=180^{\circ}$

$\displaystyle A+135^{\circ}=180^{\circ}$

$\displaystyle A=180^{\circ}-135^{\circ}$

$\displaystyle A=45^{\circ}$

Write the Pythagorean Theorem.

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

Substitute the values of the leg and hypotenuse.  The hypotenuse is the longest side of the right triangle.  Solve for the unknown variable.

$\displaystyle a^2+10^2 = 20^2$

$\displaystyle a^2 = 20^2-10^2$

$\displaystyle a^2 = 300$

$\displaystyle a=\sqrt{300} = \sqrt{100\times3} = \sqrt{100}\times \sqrt3 = 10\sqrt3$

Given that ABC is a right triangle, the length of hypotenuse BC is the root of the sum of the squares of the two other sides (in other words, $\displaystyle BC^2 = AB^2 + AC^2$. Since AB is $\displaystyle 9$ cm long and AC is $\displaystyle 5$ cm long, we get that $\displaystyle BC^2 = 81+25 = 106$, and so $\displaystyle BC = \sqrt{106}$.

When you are using Pythagorean Theorem to calculate the missing side of a right triangle, it is crucial that you identify which side is the hypotenuse, $\displaystyle c$ in the Pythagorean equation $\displaystyle a^2+b^2=c^2$. Here you're told that side AC is the longest of the three sides, so 25 will serve as the length of the hypotenuse and the value of $\displaystyle c$. This allows you to set up the equation:

$\displaystyle a^{2}+9^{2}=25^{2}$

And then you can perform the calculations on the known values:

$\displaystyle 81+a^2=625$

Meaning that:

$\displaystyle a^2= 544$$\displaystyle a^{2}=544$

From there you can simplify, arriving at a = 4 times the square root of 34.

$\displaystyle a^2= 544$$\displaystyle a^{2}$

Since the angle of the isosceles is $\displaystyle 68^\circ$, the larger angle of the right triangle formed by $\displaystyle h$ is also $\displaystyle 68^\circ$.

Using $\displaystyle tan(68)$, we can find $\displaystyle h$: 

$\displaystyle tan(68)=\frac{h}{3}$.

Then solve for $\displaystyle h$: 

$\displaystyle 3tan(68)=h$.

Simplify: $\displaystyle 7.42526056=h$.

Lastly, round and add appropriate units: $\displaystyle 7.4ft=h$.

In the diagram, AB is cut in half by the altitude. 

From here it easy to use right triangle trigonometry to solve for AC.

$\displaystyle \\ \cos 29^\circ =\frac{opp}{hyp}= \frac {5}{AC} \\* \\* AC = \frac {5}{cos 29^\circ} \approx 5.7$

The formula for the area of a triangle is:

 $\displaystyle A=\frac{1}{2}\cdot base\cdot height$

We already know what the base is and we can find the height by dividing the isosceles triangle into 2 right triangles: 

From there, we can use the Pathegorean Theorem to calculate height:

$\displaystyle 9^{2}=h^{2}+4^{2}$

$\displaystyle h^{2}=81-16=65$

$\displaystyle h=\sqrt{65}$

To find the area, now we just plug these values into the formula:

$\displaystyle A= \frac{1}{2}\cdot8 \cdot \sqrt{65}$

$\displaystyle A=32.25 cm^{2}$

The first step to solve for area is to divide the isosceles into two right triangles:

From there, we can determine the height and base needed for our area equation $\displaystyle A=\frac{1}{2}\cdot b \cdot h$ 

$\displaystyle sin(67.5^{\circ})=\frac{1/2Base}{12\; cm}$

$\displaystyle 0.924=\frac{1/2Base}{12\; cm}$

$\displaystyle 1/2Base=11.1\;cm$

$\displaystyle Base=22.2 \; cm$

From there, height can be easily determined using the Pathegorean Theorem:

$\displaystyle (12\; cm)^{2}=(11.1\; cm)^{2}+h^{2}$

$\displaystyle h^{2}=144\; cm^{2}-123.2\; cm^{2}=20.8cm^{2}$

$\displaystyle Height=4.6\; cm$

Now both values can be plugged into the Area formula:

$\displaystyle A=\frac{1}{2}\cdot22.2\; cm\cdot4.6\; cm=51.1cm^{2}$