Question number 23. Square ABCD has side length of 1. Point M is the midpoint of the side BC. Point N is the midpoint of the segment BM. Points P and Q are the points of intersection of the diagonal BD with the segments AM and AN respectively. What is the area of the shaded triangle here in green? What we can do right away is express that area as the difference of areas of several triangles. So the area we want is the area of the triangle ABM and then we subtract from that the area of the triangle ABQ and also we need to subtract the area of the triangle BMP and so we can start computing that. We'll use the formula 1 half times the base times the height and what we have is that the area of the shaded triangle for the first computation we have the triangle ABM here and we can label its dimensions. The square has side length of 1 so that's the base and then the height of that triangle is the length of the segment here BM and that is exactly 1 half because M is the midpoint of the segment BC. So that's 1 half, the base is 1 and the height is 1 half. And so next for the area of triangle ABQ we have 1 half times the base which is also 1 and the height here I will label as alpha and alpha in the picture is going to be the length of the segment over here that I just drew in so that's alpha. And finally the area of the triangle BMP, 1 half, the base is the segment BM so that's 1 half as we already found out and the height here I will label as beta and beta is the segment connecting the point P to the side of the square here BC and we demand that is the perpendicular here line connecting those and that is that is going to be beta so we can start simplifying we have 1 fourth minus 1 half alpha minus 1 fourth beta and then we can factor out a 1 fourth so 1 fourth 1 minus 2 alpha minus beta like that and now all we have to compute are the values of alpha and beta and one way we can do that is outlined in the suggested solutions so we can assign coordinates to these points so let's say that the origin is the point A then the point B will have coordinates 1,0 point D will have coordinates 0,1 and point C will have coordinates 1,1 so we can write down the equation of the line here DB the line the diagonal and that line has the equation Y is equal to negative X plus 1 so that's the equation of our line and then we know that the point M will have coordinates 1,1 half and the point N here will have coordinates 1,1 quarter and we can use the equations of the lines here the line connecting A with M will have coordinates I will have the equation Y is equal to half X and similarly Y is equal to 1 fourth X will be the equation of the line connecting A with N and so we can then compute the coordinates of point P by solving a system of equations and that is two thirds one third that's the coordinate of point P so that tells us that beta is equal to exactly one third so that's one minus two thirds and point Q here has coordinates four fifths comma one fifth so alpha is equal to one fifth and then we can finish our computation so we have one quarter one minus two times alpha so that would be two fifths minus beta which would be one third and so let me finish that over here so we have one fourth times three fifths minus one third we can cross multiply here so that's one fourth and then we have nine minus five fifteens so that gives us four fifteens times one fourth and that simplifies to 1 over 15 and that is our answer for the area and that is choice P over here 1 over 15 is the area of the shaded triangle in green

shaded triangle

square ABCD

area calculation

midpoints

coordinates