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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.

 Jan 27, 2020
 #1
avatar+18826 
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Let triangle(ABC) be an equilateral triangle with M the midpoint of side BC.

This means that AM is also an altitude of the triangle, so angle(AMB) is a right angle.

 

Draw square ALMN with diagonal AM in such a way that point N of the square is on the same side of AM as vertex B.

Call the point where MN intersects AB point X.

 

The area of the equilateral triangle(ABC) can be found by using the formula:  Area  =  ½·AB·BC·sin(60°)

--->     Area  =  ½·4·4·sin(60°)  =  6.92820323.

 

To find the area of triangle(MBX, I will need to know the length of side BX.

Since AM is a diagonal of a square, angle(AMN) = 45° and angle(BMN) = 45°.

Since B is a corner angle of an equilateral triangle, angle(B) = 60°.

Therefore, in triangle(BMX), the third angle = 75°.

Using the Law of Sines:  BX/sin(45°)  =  2/sin(75°)     --->     BX  =  1.464101615.

The Area( triangle(BMX) )  =  ½·BX·BM·sin(60°)     --->     Area( triangle(BMX) )  =  ½·1.46101615·2·sin(60°)

--->     Area  =  1.267949192

There is another triangle on the other side of the altitude of the same size; combining these two areas, we get:

--->     Area  =  2.535898385     

 

Subtracting this from the area of the triangle, we get:  6.92820323 - 2.535898385  =  4.392304845,

 (which is the area common to the square and the triangle.)

 Jan 27, 2020
 #2
avatar+519 
+1

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.

 

 Jan 28, 2020
edited by Dragan  Mar 3, 2020

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