Area of a triangle ABC is equal to 12
. On a straight line AU point taken D so
dot C is the midpoint of the segment AD. Dot K- middle side AB,
straight KD crosses side BC at the point L.
a) Prove that BL:LC=2:1.
b) Find the area of the triangle BLK.
To begin with, we will carefully make a drawing, marking the equality of the segments along the way.
Now it is easy to see that by connecting the dots AT and D, we get a triangle ABD,
wherein DK and sun are medians by definition (do you remember it?)
And the medians at the point of intersection are divided by 2: 1
counting from the top.
It is done. Write, can you prove this property yourself?
Find the area of a triangle BLK can be different. Let AE- third median
triangle ABD, it will pass through the point L intersection of the first two.
Median sun divides the triangle ABD into two equal triangles.
Therefore, the area ABD twice the area ABC and equal to 12 2 = 24.
Three medians divide the triangle into six triangles of equal area.
From here it is easy to find the area of the desired triangle BLK. 24:6 = 4
.
I note that both of these statements should also be able to prove.
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You can compare the areas of triangles BLK and ABC without touching the median.
These triangles have a common angle AT Let's use this fact.
Let's find the area ratio:
So the area BLK three times the area ABC.
The area of triangle ABC is 198. The bisector AL intersects the median BM at point K. Find the area of the quadrilateral MCLK if BL:CL=7:4 is known.
Building a sketch:
It is rather difficult to immediately see the progress of solving the problem, but we can always pose the question: what can be found using the data in the condition and the properties known to us?
We can determine the areas of some triangles, consider:
Since AM \u003d MC, then the areas of the triangles will be equal, that is:
Consider triangles ALB and ALC. The condition says BL:CL=7:4. Let's introduce the coefficient of proportionality "x" and write down the formulas for their areas:
The area ratio will be:
We also know that S ALB +S ALC =198. We can calculate the area:
Please note that we are not given any angles and linear dimensions (element lengths) in the condition, so you should not spend effort on calculating angles and lengths (sides, medians, bisectors, etc.). Why?
When the ratios of segments (angles) are given in the condition and there is not a single specific value, then most likely with such data it is possible to construct many variants of the figure. *Not for every student it is possible to see it immediately, experience is needed.
Therefore, in such cases, strive to use ratios - namely: ratios of elements, areas, use the similarity of triangles if possible.
Here we can find the ratio of the sides of the triangle. Let's express the areas of the triangles:
Based on the fact that AM=MC it follows that
Now attention! We are close to the denouement. There is another relation from which we can establish the ratio of the areas of two triangles. Express the areas of the triangles.
Let it be required to determine the area of triangle ABC. Let us draw straight lines through its vertices C and B, parallel to the sides AB and AC.
We get a parallelogram ABDC. Its area is equal to the product of the base AB and the height CO. The parallelogram ABDC consists of two equal triangles ABC and BCD, therefore, the area of the triangle ABC is equal to half the area of the parallelogram, i.e. S\(\Delta\)ABC = 1/2 AB CO.
From here: The area of a triangle is half the product of its base times its height.
S \(\Delta\) = \(\frac(a h)(2)\)
This formula can be represented as follows:
S \(\Delta\) = \(\frac(a)(2)\) h, or S \(\Delta\) = a\(\frac(h)(2)\).
Formulas for calculating the area of a triangle
1. From geometry, the Heron formula is known:
$$ S = \sqrt(p (p - a)(p - b) (p - c)),$$
(where p = ( a + b + c) / 2 - semi-perimeter), which allows you to calculate the area of a triangle on its sides.
2 . Theorem. The area of a triangle is equal to half the product of two sides and the sine of the angle between them:
S=1/2 bc sinA.
Proof. It is known from geometry that the area of a triangle is equal to half the product of the side of the triangle and the height dropped to this side from the opposite vertex.
S=1/2 b h b (1)
If angle A is acute, then from the triangle ABH we find BH = h b = c sinA.
If angle A is obtuse, then
HH = h b = c sin (π - A)= With sinA.
If angle A is right, then sin A = 1 and
hb=AB= With = With sinA.
Therefore, in all cases h b = c sin A. Substituting into equality (1), we obtain the formula to be proved.
In the same way, we get the formulas: S = 1 / 2 ab sin C= 1 / 2 ac sin B
3. Based on the sine theorem:
$$ b = \frac(a sinB)(sinA); \;\; c = \frac(a sinC)(sinA) $$
Substituting these expressions into formula (1), we obtain the following formula:
$$ S = \frac(a^2 sinB sinC)(2sinA) $$
