Triangle - Examples, Exercises and Solutions

We add the three angles to see if they equal 180 degrees:

$30+60+90=180$
The sum of the angles equals 180, so they can form a triangle.

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

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

We add the three angles to see if they are equal to 180 degrees:

$56+89+17=162$

The sum of the given angles is not equal to 180, so they cannot form a triangle.

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:

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

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

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

The formula to calculate the area of a triangle is:

(side * height corresponding to the side) / 2

Note that in the triangle provided to us, we have the length of the side but not the height.

That is, we do not have enough data to perform the calculation.

We add the three angles to see if they are equal to 180 degrees:

$90+115+35=240$
The sum of the given angles is not equal to 180, so they cannot form a triangle.

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

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:

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

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:

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

The formula to calculate the area of a triangle is:

(side * height descending from the side) /2

We place the data we have into the formula to find X:

$20=\frac{AB\times AC}{2}$

$20=\frac{x\times5}{2}$

Multiply by 2 to get rid of the fraction:

$5x=40$

Divide both sections by 5:

$\frac{5x}{5}=\frac{40}{5}$

$x=8$

Since in the given triangle all angles are equal, all sides are also equal.

It is known that in an equilateral triangle the measure of the angles will always be equal to 60° since the sum of the angles in a triangle is 180 degrees:

$60+60+60=180$

Therefore, it is an equilateral triangle.

The perimeter of the triangle is equal to the sum of all sides together, therefore:

$6+8+10=14+10=24$

The perimeter of a triangle is equal to the sum of all its sides together:

$11+7+13=11+20=31$