ABC is an equilateral triangle with side length 4. M is the midpoint of BC, and AM is a diagonal of square ALMN. Find the area of the region common to both ABC and ALMN.
To find the area of the region common to both ABC and ALMN, we need to calculate the area of the smaller triangle formed by the common region.
Let's start by drawing both triangles ABC and ALMN:
A
/ \
/ \
/_____\
B C
L ___ N
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|_____|
We know that ABC is an equilateral triangle with side length 4. This means that the height of the triangle is the length of one of its sides times the square root of 3 divided by 2. In this case, the height of ABC is 4√3/2 = 2√3.
Now, let's find the area of triangle ABC. The area of an equilateral triangle is given by the formula:
Area = (side length)^2 * sqrt(3) / 4
Since the side length of ABC is 4, the area of ABC is:
Area_ABC = 4^2 * sqrt(3) / 4
= 16 * sqrt(3) / 4
= 4 * sqrt(3)
Next, let's find the area of triangle ALM. Since AM is a diagonal of square ALMN and the side length of the equilateral triangle ABC is 4, the diagonal AM is also 4. This means that the side length of the square ALMN is 4.
The area of a square is given by the formula:
Area = (side length)^2
So, the area of square ALMN is:
Area_ALMN = 4^2
= 16
The area of the smaller triangle formed by the common region (triangle ALM) is half the area of the square ALMN because AM is the diagonal of the square and the base of the smaller triangle. Therefore, the area of triangle ALM is:
Area_ALM = Area_ALMN / 2
= 16 / 2
= 8
Finally, the area of the region common to both ABC and ALMN is the area of triangle ALM, which is:
Area_common = Area_ALM
= 8
So, the area of the region common to both ABC and ALMN is 8.