Area of Isosceles Triangles

The triangle we are looking at is the large triangle - ABC

The triangle is formed by three sides AB, BC, and CA.

Now let's remember what we need for the calculation of a triangular area:

(side x the height that descends from the side)/2

Therefore, the first thing we must find is a suitable height and side.

We are given the side AC, but there is no descending height, so it is not useful to us.

The side AB is not given,

And so we are left with the side BC, which is given.

From the side BC descends the height AD (the two form a 90-degree angle).

It can be argued that BC is also a height, but if we delve deeper it seems that CD can be a height in the triangle ADC,

and BD is a height in the triangle ADB (both are the sides of a right triangle, therefore they are the height and the side).

As we do not know if the triangle is isosceles or not, it is also not possible to know if CD=DB, or what their ratio is, and this theory fails.

Let's remember again the formula for triangular area and replace the data we have in the formula:

(side* the height that descends from the side)/2

Now we replace the existing data in this formula:

$\frac{CB\times AD}{2}$

$\frac{11.6\times3}{2}$

$\frac{34.8}{2}=17.4$

This question is a bit confusing. We need start by identifying which parts of the data are relevant to us.

Remember the formula for the area of a triangle:

The height is a straight line that comes out of an angle and forms a right angle with the opposite side.

In the drawing we have a height of 6.

It goes down to the opposite side whose length is 5.

And therefore, these are the data points that we will use.

We replace in the formula:

$\frac{6\times5}{2}=\frac{30}{2}=15$

First, we will identify the data points we need to be able to find the area of the triangle.

the formula for the area of the triangle: height*opposite side / 2

Since it is a right triangle, we know that the straight sides are actually also the heights between each other, that is, the side that measures 5 and the side that measures 7.

We multiply the legs and divide by 2

$\frac{5\times7}{2}=\frac{35}{2}=17.5$

First, let's remember the formula for the area of a triangle:

(the side * the height that descends to the side) /2

In the question, we have three pieces of data, but one of them is redundant!

We only have one height, the line that forms a 90-degree angle - AD,

The side to which the height descends is CB,

Therefore, we can use them in our calculation:

$\frac{CB\times AD}{2}$

$\frac{8\times9}{2}=\frac{72}{2}=36$

Due to the fact that AB is perpendicular to BC and forms a 90-degree angle,

it can be argued that AB is the height of the triangle.

Hence we can calculate the area as follows:

$\frac{AB\times BC}{2}=\frac{8\times6}{2}=\frac{48}{2}=24$