document.write( "Question 1151273: Equilateral triangle △ABC has side length 2, M is the midpoint of
\n" );
document.write( "segment AC, and C is the midpoint of BD. What is the area of
\n" );
document.write( "△CDM?
\n" );
document.write( "(please check the link below as it contains the image for the question)
\n" );
document.write( "https://i.imgur.com/g3InsY0.png \n" );
document.write( "
Algebra.Com's Answer #772975 by jim_thompson5910(35256) ![]() You can put this solution on YOUR website! \n" ); document.write( "Let's explore hint #2 that the tutor @ikleyn has provided.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Triangle ABC has base of BC = 2 \n" ); document.write( "Let h be the height of triangle ABC. It does not matter what h is for this thought experiment. \n" ); document.write( "The area of triangle ABC is therefore, A = (1/2)*base*height = (1/2)*2*h = h. \n" ); document.write( "Area of triangle ABC = h.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The base of triangle CDM is also 2, because C is the midpoint of BD, so BC = CD = 2. \n" ); document.write( "The height of point M is half that of point A's height. Imagine that point A is (0, h). Through use of the midpoint formula, you'll find that the y coordinate of point M will be y = h/2. \n" ); document.write( "So the height of triangle CDM is h/2.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "area of triangle CDM = (1/2)*base*height \n" ); document.write( "area of triangle CDM = (1/2)*2*(h/2) \n" ); document.write( "area of triangle CDM = h/2 \n" ); document.write( "area of triangle CDM = (area of triangle ABC)/2 \n" ); document.write( " \n" ); document.write( " |