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.
I drew the diagram but I don't know how to find the answer? I think it has something to do with the law of sines and cosines
Help is appreciated, thanks
I started with AB, the vertical diagonal of the square.
I then drew square ALMN
At M, I drew the base BC of the equilateral triangle perpendicular to AM, letting BM = MC = 2
Joi AB and AC for the triangle.
Let the intersection of ML and AB be P, and
the intersection of MN and AC be Q
clearly triangles APM and AQM are congruent and form the overlap, or common area
So all we need is the area of triangle APM, then double that.
In triangle APM, angle PMA = 45°, anglePAM = 30°
then angle APM = 105°
Also it is easy to show that AM = 2√3
let's find PM
PM/sin30 = 2√3/sin 105
PM = 2√3(sin30)/sin105 = √3/sin105
area of triangle APM = (1/2)(AM)(PM)sin45
= (1/2)(2√3)(√3/sin105)(√2/2)
= (6/sin105)(√2/2)
= 3√2/(2sin105)
I got appr 2.196
better check my algebra and arithmetic, should have written it out on paper first.
If you need an "exact" answer,
change sin(105)
= sin(60+45)
= sin60cos45 + cos60sin45
= ....
then sub back into above answer.
Thanks a lot
To find the area of the region common to both triangle ABC and square ALMN, we can break it down into two parts: the shaded region within the equilateral triangle and the shaded region within the square.
Let's start by finding the area of the shaded region within the equilateral triangle ABC.
Since ABC is an equilateral triangle with side length 4, we can draw the altitude from A to BC, which meets BC at point D. This altitude divides ABC into two congruent 30-60-90 right triangles.
Using the properties of a 30-60-90 triangle, we can determine that the length of the altitude AD is equal to 4 * √3 / 2 = 2√3.
Now, let's focus on the shaded region within square ALMN.
Since AM is a diagonal of the square and the side length of the square is unknown, let's calculate it. We know that AM is the diagonal of square ALMN and the length of AM is equal to the altitude AD of triangle ABC. Thus, the length of the side of square ALMN is also 2√3.
Now, to find the area of the shaded region within square ALMN, we need to find the area of the square and subtract the non-shaded areas within the square.
The area of square ALMN with side length 2√3 is (2√3)² = 4 * 3 = 12.
Now, let's subtract the non-shaded areas within the square.
Since ABC is an equilateral triangle and AM is a diagonal of square ALMN, we can see that triangle ABC and square ALMN are both congruent. Therefore, the area of triangle ABC is equal to the area of square ALMN.
So, the area of the shaded region within square ALMN is the same as the area of triangle ABC, which is (1/2) * base * height = (1/2) * 4 * 2√3 = 4√3.
Now, let's find the area of the region common to both ABC and ALMN by adding the areas of the two shaded regions:
Area of the region common to both ABC and ALMN = Area of the shaded region within the equilateral triangle + Area of the shaded region within the square = 4√3 + 4√3 = 8√3.
Therefore, the area of the region common to both ABC and ALMN is 8√3.
To find the area of the region common to both triangles ABC and ALMN, we need to calculate the areas of these two triangles separately and then find the overlapping area.
First, let's find the area of equilateral triangle ABC. Since ABC is equilateral with a side length of 4, we can use the formula for the area of an equilateral triangle, which is `Area = (sqrt(3) / 4) * side^2`. Plugging in the values, we get:
Area(ABC) = (sqrt(3) / 4) * 4^2 = (sqrt(3)/4) * 16 = 4sqrt(3).
Next, we need to find the area of square ALMN. Since AM is a diagonal, we know that the side length of the square is equal to the hypotenuse of right triangle AMN (45-45-90 triangle). Given that triangle AMN is isosceles, we can use the Pythagorean theorem to find the side length of the square:
AM = sqrt(AB^2 + BM^2) [By Pythagorean theorem]
= sqrt(4^2 + (4/2)^2)
= sqrt(16 + 4)
= sqrt(20)
= 2sqrt(5).
Now that we have the side length of the square, we can find its area:
Area(ALMN) = side^2 = (2sqrt(5))^2 = 4*5 = 20.
Now, let's calculate the overlapping area. The area common to both triangles can be found by subtracting the area of the unshared parts of the square from the area of the equilateral triangle.
The unshared parts of the square consist of two right-angled triangles, MNK and ALK. Each of these triangles is a 45-45-90 triangle, and we can calculate their areas using the formula: `Area = (1/2) * base * height`.
Since both base and height are equal in these two triangles, we can consider the length of one side of each triangle as the base, resulting in:
Area(MNK) = Area(ALK) = (1/2) * x * x,
where x represents the length of one side of the square.
Summing up the areas of MNK and ALK, we get:
Total area unshared = 2 * (1/2) * x * x = x^2.
Since we know that the side length of the square is 2sqrt(5), we can plug this into the equation:
Total area unshared = (2sqrt(5))^2 = 4 * 5 = 20.
Finally, we can find the area of the region common to both triangles by subtracting the unshared area from the area of the equilateral triangle:
Area common = Area(ABC) - Total area unshared
= 4sqrt(3) - 20.
Therefore, the area of the region common to both triangles ABC and ALMN is 4sqrt(3) - 20.