Area of Isosceles Triangles - Examples, Exercises and Solutions

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:

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

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:

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

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

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

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

The formula for calculating the area of a triangle is:

(the side * the height from the side down to the base) /2

That is:

$\frac{BC\times AE}{2}$

We insert the existing data as shown below:

$\frac{4\times5}{2}=\frac{20}{2}=10$

The formula for the area of a triangle is

$A = \frac{h\cdot base}{2}$

Let's insert the available data into the formula:

(7*6)/2 =

42/2 =

21