Welcome to the video on solving area problems with triangles and trapezoids.
During this video we will work through detailed exercises that apply geometry and trigonometry.
These concepts are fundamental to architecture, engineering and urban design.
Let's get started!
This video covers.
First.
Exercises on triangles, where we will see the calculation of
- Base areas.
- Calculating heights from trigonometric ratios.
- And applications to figures such as the pentagon, where its geometry can be broken down into triangles.
Second.
Exercises on trapezoids, where we will see the calculation of
- Areas in different situations: known sides, diagonals and equivalent methods.
We are going to start with triangle problems.
We will look at different ways of calculating area and height
depending on the data available: base and height, three sides or trigonometric ratios.
These methods will allow us to deal with any situation.
In this slide, we have a triangle for which we want to calculate the
area. As you know, the area of a triangle is determined by applying the formula base times height, divided by two.
In our case, the base, which we will denote as c, measures 50 metres.
The height, which we have determined, is 26.66 metres.
Substituting these values in the formula, and performing the
operations, we have that the approximate area of this triangle is 666.58 square metres.
This calculation provides a clear understanding of the figure's area,
using a straightforward process: identify the base, measure the
perpendicular height and apply the formula. In the following
example, we will see how the situation changes when we change the data.
In this second case, and as we can see in the picture, we are going
to apply Heron's formula to calculate the area of a triangle when we
know its three sides. This formula is especially useful when we don't have the height directly.
First, we identify our data:
Side a = 45 metres.
Side b = 30 metres.
Side c = 50 metres.
With these three sides, we calculate the semi-perimeter, where s is half of the total perimeter.
Thus, s is equal to 62.5 metres.
Once we have "s", we apply Heron's formula for the area, where
substituting the values and performing the operations, we obtain that the approximate area of this triangle is 666.58 square metres.
Notice that we arrive at the same area as with the base-height method
of the previous exercise, which serves as a check. This method is
highly versatile because it does not rely on knowing the height in
advance, facilitating the calculation in cases where we only have the lengths of the sides.
Now, let's look at a third way to calculate the area of a triangle:
when we know two of its sides and the angle between them. This is a
very useful application, for example, in situations where it is not easy to measure the height directly.
In the first case, we have:
Side b = 30 metres.
Base c = 50 metres.
Angle Alpha: 62.72 degrees
The formula to be used is the same as in the first case (Base-Height
Formula), where the height will be equal to side b times the sine of angle alpha.
Substituting the values and performing the operations, we obtain that
the approximate area of this triangle is 666.58 square metres. In the second case, the solution is analogous.
It is noteworthy that, irrespective of the method applied (either
with base and height, with Heron's formula, or with the information
of two sides and the angle between them), we arrive at a very
similar area. This consistency between different methods is an excellent way of verifying the correctness of our result.
Another alternative method of calculating the area of a triangle, is
when we know the three lengths of its sides and in addition the radius of the circumscribed circle, often denoted as Rcc.
In this case we have:
Side a = 45 metres.
Side b = 30 metres.
Base c = 50 metres.
Radius of the circumscribed circumference equals 25.32 metres.
The formula is equal to the product of the three sides divided by
four times the radius of the circumscribed circle. Substituting the
values, we see, once again, that the result is consistent with the areas obtained by the previous methods.
This consistency is an excellent indicator that our calculation is correct.
Finally, we will look at an alternative method for calculating the
area of a triangle when we know the semi-perimeter, which we will
call s, and the radius of the inscribed circle, which we will call R I c.
First, let's remember that the semi-perimeter ‘s’ is calculated from the sum of the three sides divided by two.
In our example, the sides are:
Side a = 45 metres.
Side b = 30 metres.
Base c = 50 metres.
Radius of the inscribed circle equal to 10.67 metres.
Therefore the semi-perimeter s will be equal to 62.50 metres.
Knowing s and the radius of the inscribed circle, the formula for the area will be their product.
Substituting the values and making the product, we obtain that the approximate area of this triangle is 666.58 square metres.
In summary, we explored multiple approaches to calculate the same triangle’s area:
- Using base and height directly.
- Applying Heron's formula, knowing the three sides.
- Using two sides and the angle between them.
- Using the three sides and the radius of the circumscribed circle.
- Using the three sides and the radius of the inscribed circle.
All these methods lead to the same result, which gives us solid confidence in the accuracy of the calculated area.
With this, we conclude the different ways of finding the area of a triangle.
In this exercise, the objective is to find the height of a tower using two measurements taken from positions A and B.
First, from a point further away from the tower (A), we measure an
angle of elevation of 25 degrees. Then, we move 89.78 metres
towards the tower, standing at point B, from where we again measure the angle, this time obtaining 42 degrees.
The idea is simple and we can solve it from the following steps:
1.
We call x the distance from the tower to the nearest point B.
2.
Then, the distance from the tower to the furthest point A is x + 89.78 metres.
3.
With the trigonometric ratios (the tangent of the angle is the
quotient between the height of the tower and the horizontal
distance), we establish two equations: one from the angle from A and the other from B, and we clear the height h.
4.
Equating both expressions describing the height of the tower, we can solve for x.
5.
Once x has been calculated, we substitute that value into either equation to find the height h.
These calculations indicate that the tower is approximately 86.84 metres tall,
demonstrating the practical application of trigonometric methods.
This method combines angular measurements and the distance
between the two positions to deduce the height without the need to climb the tower or have a direct measurement.
In this exercise, you want to determine the total height of a tower (O M) by looking at it from a single viewpoint V.
From V, two angles are measured: one of elevation towards the top
(25 degrees) and one of depression towards the base (42 degrees).
In addition, the distance from the viewpoint V to the base of the tower O is known to be 85 metres.
We can solve the exercise from the following steps:
1.
Using the angle of depression (42 degrees), and the distance to the
base (85 metres.), we calculate how much the base of the tower
descends with respect to point V. This gives us the height MO, which is the portion of the tower below the level of point V.
2.
Once we know MO, we determine the horizontal distance VM, which is
the projection of the line of sight to the base of the tower (VO) on
a horizontal line at the height of V, obtaining the magnitude sought.
3.
With VM and the angle of elevation (25 degrees), we find MZ, which is the height of the tower above point V.
Finally, we add the two heights: MO (bottom) plus MZ (top) to obtain the total height of the tower OZ.
After doing these steps, the total height of the tower is approximately 86.33 metres.
In this case, we are going to calculate the side and area of a regular pentagon inscribed in a circle of radius 25 metres.
The starting point is to imagine the pentagon inside the circle.
By drawing lines from the centre of the circle to each of the five vertices of the pentagon,
the pentagon is divided into five congruent triangles (equal in shape and size).
If we consider one of these triangles, for example AOB,
we will notice that the angle at the
centre, the angle AOB, is the result of dividing the 360 degrees of the circumference by 5, since the pentagon has five vertices.
This gives us a central angle of 72 degrees.
Next, we draw the apothem of the pentagon, which is the line from the
centre to the midpoint of one of the sides, say OM.
This apothem divides the triangle AOB into two smaller, congruent right triangles: AOM and B O M.
By doing this, each of these smaller triangles has an angle at O of 36 degrees (half of 72 degrees).
Now, in one of these right triangles, say A O M, we know the
hypotenuse (the radius, 25 metres) and the angle of 36 degrees.
By applying trigonometric ratios, we can determine half the side of the pentagon.
Since AM is half the side, it will be equal to 25 times the sine of 36 degrees.
This gives us AM approximately equal to 14.69 metres.
Since AM is half the side, the whole side AB of the pentagon will be twice as long, or about 29.39 metres.
Next, to calculate the apothem, we use the cosine of 36 degrees,
so OM equals 25 times the cosine of 36 degrees, which gives us about 20.23 metres for the apothem.
With this data, we can now calculate the area of the pentagon.
Recall that the area of the pentagon is five times the area of one of the triangles that make up the pentagon.
To obtain the area of one of these triangles,
we use directly the complete base AB and the height O M.
In this case, we take triangle AOB, which has a base AB of 29.39
metres and a height O M of 20.23 metres. The area of this triangle is a half of the base times the height.
That is: approximately 297.2 square metres.
Since the pentagon has five of these triangles, we multiply 297.2 square metres by 5,
giving a total area of the pentagon of approximately 1,486.02 square metres.
In short, we have found the side of the pentagon (approximately 29.39 metres)
and its total area (about 1,486 square metres) from the
radius of the circumference that circumscribes it, using only basic geometry and trigonometry concepts.
We now move on to trapezoid exercises, where we will calculate the
area in various situations using data such as bases, height, sides,
diagonals or angles. These methods are very useful in engineering and architectural applications.
In this first example related to trapezoids, we have the section of a
structural portico of a sports hall in the shape of a trapezoid.
The objective is to calculate the area of that section, knowing the lengths of its major base, minor base and height.
The data are as follows:
Larger base (c) equal to 50 metres.
Minor base (a) equal to 20 metres.
Height (h) equal to 24 metres.
One of the simplest formulas for finding the area of a trapezoid is
to first calculate the so-called ‘median base’, which is obtained by
adding the major base and the minor base, and dividing that result by two.
Substituting the values gives us an median base equal to 35 metres.
Once the median base has been obtained, the area of the trapezoid is found by multiplying this median base by the height h:
In this way, the area of the trapezoid-shaped portico section turns out to be 840 square metres.
In this second case, we want to calculate the area of a trapezoid
from the previous case but knowing its four sides: the major base,
the minor base and the two oblique sides. Unlike the previous case,
here we do not have the height directly, so we must deduce it from the available data.
The initial data are:
Minor base (a) equal to 20 metres.
Oblique side (b) equal to 25 metres.
Larger base (c) equal to 50 metres.
Oblique side (d) equal to 33.24 metres.
The key to finding the height h is to decompose the trapezium into
simpler figures (a rectangle and two right triangles) and apply the Pythagorean theorem.
Although the mathematical process is more
extensive than in the previous case, the end result of this calculation is that the height h of the trapezoid is 24 metres.
We can also determine the height of the trapezoid by directly applying the formula shown in the slide.
With the height determined, we return to the basic formula for the area of a trapezoid,
which is equal to the sum of the bases (a + c), divided by two and multiplied by the height.
Substituting the values we are left with the resulting area equal to 840 square metres.
Once again, we obtain an area of 840 square metres, now calculated
from the four sides of the trapezoid and using trigonometry and the Pythagorean theorem to deduce the height.
In this third case, the challenge is to calculate the area of the
same trapezoid knowing the lengths of its diagonals and the angle between them.
This approach uses a geometric transformation that
simplifies the trapezoid problem into finding the area of an equivalent triangle.
The data we start with are:
Length of the first diagonal (dg1) equal to 36.12 metres.
Length of the second diagonal (dg2) equal to 49.24 metres.
Angle between the diagonals (Theta) equal to 70.80 degrees.
(Remember that the supplementary angle (Epsilon) is 109.20 degrees,
but the important thing is that the sine of (Theta) and the sine of (Epsilon) will give us the same height).
First, by a simple change of position (moving the diagonal BD to
vertex C, the trapezium is transformed into an equivalent triangle (ACQ).
Now, knowing the length of the diagonals and the angle they form, we can determine the height h of the triangle:
The height h is obtained by multiplying the length of one of the
diagonals by the sine of the angle between them. Substituting the values we have that h is equal to 46.50 metres.
Once the height h is obtained, the area of the equivalent triangle
ACQ is simply half the product of the other diagonal (dg1) and the height h, resulting in an area of 840 square metres.
The most interesting thing about this method is that we start from less common data
(diagonals and angle between them) and, by means of a simple geometric transformation, we arrive at a coherent result.
In this example, we start again from the same initial trapezoid
of which we know its major base, its minor base and its height.
The data are the same as in the first exercise of this chapter on trapezoids:
Major base (c) equal to 50 metres.
Minor base (a) equal to 20 metres.
Height (h) equal to 24 metres.
As we have already seen, the area of the trapezoid can be easily calculated:
First, we find the mean base (MN), which is the average of the major base and the minor base.
Thus MN will be equal to 35 metres.
With the product of the average base and the height, we get the area of the trapezoid equal to 840 square metres.
So far, it is the same result we got before.
But now, this exercise shows us how the area of the trapezoid
is equivalent to the area of a rectangle with the same dimensions of average base and height,
and also to the area of a square built from the ratio between the average base and the height.
The idea is as follows:
We start with the original trapezium, ABCD.
Through a process of geometric equivalences (such as shifting areas
and extending or reducing lines), we can transform it into a
rectangle A'B'C'D'(prime), that has the same base and height dimensions as the average base and height of the trapezoid.
Therefore, rectangle A'B'C'D'(prime) also has an area of 840 square metres.
Finally, we can form a square A’’B’’C’’D’’(double prime) whose area
is equal to that of the trapezium (and the rectangle).
To achievethis, the side of the square must be equal to the square root of the product of the average base times the height.
Calculating the side L gives us approximately 28.98 metres.
The area of this square is therefore L square which will be approximately 840 square metres.
Thus, it is shown that the area of our trapezium can be expressed
in different equivalent ways: as that of a rectangle with the
appropriate dimensions or as that of a square constructed by geometric relations.
The second part of the exercise goes one step further,
using the Pythagorean Theorem in an additional construction, which leads us to another equivalent situation.
Thus, we also obtain a rectangle A’B’X’Z’(prime)
with a given area, and then a square Z’’B’’J’’K’’(double prime) with identical area.
This square has a side L’ (prime)
calculated from another smaller height h’ (prime) (11
metres in the example), giving a smaller area (385 square metres),
which confirms the Pythagorean relations and the internal coherence of the whole.
In summary, this exercise illustrates how, starting from the
average base of the trapezoid and its height,
it is possible to demonstrate very interesting geometric equivalences, obtaining in a
clear way that the area of the trapezoid is exactly the same as that of a rectangle or a square designed from the same key dimensions.
Thank you for joining me on this journey through the fascinating world of triangles and trapezoids.
Through these slides, we’ve discovered that behind every geometric
figure lies not just definitions and formulas, but profound concepts
that allow us to better understand the reality that surrounds us.
I hope this video has been useful and interesting, and that you can now
face both the practical applications and the more theoretical
challenges that geometry presents with greater confidence and curiosity.