In a triangle, segments AB and AC are trisected, and D is the midpoint of BC. If the area of Triangle ABC is 630 cm^2, then the area of the section marked x, in cm^2, is?
Well, well, well, we have a geometry question! Triangle ABC and its secrets, let's see what we can make of it.
Since segments AB and AC are trisected, that means they are divided into three equal parts. So, let's label the points where the trisection occurs as E and F. So now we have AE, EB, AF, and FC.
Now, since D is the midpoint of BC, that means that AD is a median of the triangle, dividing it into two equal parts. So, the area of Triangle ADF will be equal to the area of Triangle ABF.
So what's left is to find the area of Triangle ADF. We can find it using the ratio of the sides AF:AD, which is 2:1.
Given that the area of Triangle ABC is 630 cm^2, the area of Triangle ADF would be 630 cm^2 divided by 3, since the ratio of the sides is 2:1.
Therefore, the area of Triangle ADF is 210 cm^2. So, the area of the section marked x would be 210 cm^2.
Voila! Mystery solved with a touch of math and a sprinkle of humor!
To find the area of the section marked "x", we can use the concept of similar triangles.
Step 1: Label the point where the trisected segments intersect the sides of the triangle.
Let's call the point where AB is trisected as E, and the point where AC is trisected as F.
Step 2: Identify the similar triangles.
In this case, triangles ABC and AEF are similar. This is because angle A is common to both triangles and the segments AE, CE, and EF are parallel to the sides of triangle ABC.
Step 3: Determine the ratio of the corresponding sides.
Since triangle ABC and triangle AEF are similar, the ratio of their corresponding sides is the same. We see that AE is one-third of AB and CF is one-third of AC. Therefore, the ratio of the corresponding sides is 1:3.
Step 4: Calculate the area of triangle AEF.
The ratio of the corresponding sides is 1:3, which means the ratio of their areas is 1^2:3^2 = 1:9. So, the area of triangle AEF is 1/9 of the area of triangle ABC.
Area of triangle AEF = 1/9 * 630 cm^2 = 70 cm^2.
Step 5: Calculate the area of the section marked "x".
The area of the section marked "x" can be found by subtracting the area of triangle AEF from the area of triangle ABC.
Area of section x = Area of triangle ABC - Area of triangle AEF
= 630 cm^2 - 70 cm^2
= 560 cm^2.
Therefore, the area of the section marked "x" is 560 cm^2.
To find the area of the section marked "x" in Triangle ABC, we can use the concept of similar triangles.
Here's a step-by-step explanation:
1. Draw a diagram of Triangle ABC, segment AD, and the section marked "x."
2. Since segment AB and AC are trisected, each segment is divided into three equal parts. Let's label the points where AB is trisected as E and F, and where AC is trisected as G and H, respectively.
3. Now, we have three smaller triangles: Triangle ADE, Triangle ABF, and Triangle ACG.
4. Notice that Triangle ADE is similar to Triangle ABC because they share an angle at A. Similarly, Triangle ABF is similar to Triangle ABC, and Triangle ACG is similar to Triangle ABC.
5. Since Triangle ADE, Triangle ABF, and Triangle ACG are all similar to Triangle ABC, their corresponding sides are proportional.
6. Let's denote the length of AD as y. Since point D is the midpoint of BC, the length of BD and CD would also be y.
7. Since segment AB and AC are trisected, each smaller segment would be y/3.
8. Now, let's determine the length of AE, EF, and AF. AE would be (2/3)y since it is 2 parts of AB, and EF would be (1/3)y since it is 1 part of AB. Similarly, AF would also be (1/3)y.
9. Using the ratio of similarity between Triangle ADE and Triangle ABC, we can write the following equation:
(Area of Triangle ADE) / (Area of Triangle ABC) = (AE^2) / (AB^2)
We know that the area of Triangle ADE = (1/2) * AD * AE and the area of Triangle ABC = (1/2) * AB * AC.
Substituting the values, we get:
[(1/2) * AD * AE] / [(1/2) * AB * AC] = [(2/3)y * (2/3)y] / [AB * AC]
Simplifying the equation, we have:
AE^2 = (4/9) * AB * AC
Since AB = 3(AF) and AC = 3(AG), we can substitute these values into the equation, giving:
(2/3)y * (2/3)y = (4/9) * 3(AF) * 3(AG)
Simplifying further, we get:
(4/9)y^2 = 4(AF)(AG)
10. Since segment AB and AC are trisected, AF and AG are also (1/3)y.
Substituting these values into the equation, we have:
(4/9)y^2 = 4((1/3)y)((1/3)y)
Simplifying, we get:
(4/9)y^2 = (4/9)y^2
11. This tells us that the area of Triangle ADE is equal to the area of Triangle ABC, but one-third of its length and width. So, the area of Triangle ADE is (1/3) * 630 cm^2 = 210 cm^2.
12. Finally, the area of the section marked "x" is equal to the area of Triangle ABC minus the area of Triangle ADE:
Area of Section x = 630 cm^2 - 210 cm^2 = 420 cm^2.
Therefore, the area of the section marked "x" in cm^2 is 420 cm^2.