Any triangle with a right angle in it is called a right triangle or a right angled triangle.

The**Pythagoras Theorem** states that in any right triangle with sides of length a, b, and c,

if the angle between sides of lengths a and b is the right angle, then $a^2+b^2=c^2$

There are two special right triangles we need to know more about for the test. These are the 30-60-90 and 45-45-90 triangles respectively.

These, with their respective special properties, are illustrated below. These properties are important to know.

Since the properties of generic triangles still hold, area of right triangles is still given by $1/2$*base*height.

However, in these cases, one of the perpendicular sides qualifies as a base, and the other as a height, so computing areas is much easier than with general triangles.

For instance, in each of the previous two figures, Area(ACB)=/2$ sq. units.

E.g. A 30-60-90 triangle X, and a 45-45-90 triangle Y, have the same length of their smallest side. The ratio of the areas of the two triangles is?

X has sides $s,sÃ3,2s$, while Y has sides $s,s,sÃ2$, if the smallest sides are of the same length.

The areas are therefore: Area(X)=0/2={s^2Ã3}/2$ and Area(Y)=$s^2/2$. Area(X)/Area(Y)=$Ã3$.

The

if the angle between sides of lengths a and b is the right angle, then $a^2+b^2=c^2$

There are two special right triangles we need to know more about for the test. These are the 30-60-90 and 45-45-90 triangles respectively.

These, with their respective special properties, are illustrated below. These properties are important to know.

Since the properties of generic triangles still hold, area of right triangles is still given by $1/2$*base*height.

However, in these cases, one of the perpendicular sides qualifies as a base, and the other as a height, so computing areas is much easier than with general triangles.

For instance, in each of the previous two figures, Area(ACB)=/2$ sq. units.

E.g. A 30-60-90 triangle X, and a 45-45-90 triangle Y, have the same length of their smallest side. The ratio of the areas of the two triangles is?

X has sides $s,sÃ3,2s$, while Y has sides $s,s,sÃ2$, if the smallest sides are of the same length.

The areas are therefore: Area(X)=0/2={s^2Ã3}/2$ and Area(Y)=$s^2/2$. Area(X)/Area(Y)=$Ã3$.

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