+63 Points D, E, and are the midpoints of sides BC, CA, and AB, respectively, of triangle ABC. Points X, Y, and Z are the midpoints of EF, FD, and DE, respectively. If the area of triangle XYZ is 21, then what is the area of triangle CXY?
+128406
I have down-scaled the problem a little, but the same logic would apply
The image below lays out the particulars :
The base of triangle XYZ = XY = 6 and the height = 6...so...the area =
6^2 / 2 = 18 sq units
Now...X = (12, 12) and Y = (15,6)
And we can call segment XY the base of triangle CXY...and this is
sqrt [ (15 -12)^2 + ( 12 - 6)^2 ] = sqrt [ 3^2 + 6^2] = sqrt (45) = 3sqrt (5) units in length = (1)
And the equation of the line passing through these two points is given by
y = -2(x -12) + 12 ... y = -2x + 36 or 2x + y - 36 = 0 = AX + By + C = 0
Now ....the distance from C to this line is the altitude of CXY and it is given by
altitude = abs (Ax + By + C) / sqrt (A^2 + B^2) where (x,y) are the coordinates of C = (0,0)
So we have that
altitude = abs ( 2(0) + 1(0) - 36 ) / sqrt ( 2^2 + 1^2) = 36 / sqrt (5) = (2)
So.....the area of triangle CXY = base * altitude / 2 = (1) * (2) / 2 =
[ 3 sqrt(5) * 36 ] / [ 2 * sqrt (5) ] = 3 * 36 / 2 = 108 / 2 = 54 sq units
Now.....since in the original problem the area of XYZ is specified as 21 sq units....and the area of XYZ triangle in the diagram = 18 sq units...we can use a proportion to find the true area of CXY....so we have....
21 / 18 = true area of CXY / 54
7 / 6 = true area of CXY / 54
true area of CXY = 7 * 54 / 6 = 7 * 9 = 63 sq units
+26367 Points D, E, and F are the midpoints of sides BC, CA, and AB, respectively, of triangle ABC.
Points X, Y, and Z are the midpoints of EF, FD, and DE, respectively.
If the area of triangle XYZ is 21, then what is the area of triangle CXY?
Areas = ?
\(\begin{array}{|rclcrcl|} \hline \triangle ABC &=& A \\ \triangle DEF &=& \frac14 A \\ \triangle XYZ &=& \frac14 \triangle DEF \\ &=& \frac14 \cdot \frac14 A \\ &=& \frac{1}{16} A = 21 \\ \frac{1}{16} A &=& 21 \\ A &=& 16\cdot 21 \\\\ \triangle CXY &=& \frac{AB}{4} \cdot \frac{h}{2} \quad & | & \sin(A) &=& \frac{h}{ \frac{AC}{2} +\frac{AC}{4} } \\ & & & | & \sin(A) &=& \frac{h}{ \frac34 AC } \\ & & & | & h &=& \frac34 AC \sin(A) \\ \triangle CXY &=& \frac{AB}{4} \cdot \frac{ \frac34 AC \sin(A) }{2} \\ &=& \frac{3}{32} \cdot AB\cdot AC\cdot \sin(A) \quad & | & AB\cdot AC\cdot \sin(A) &=& 2A \\ &=& \frac{3}{32} \cdot 2A \\ &=& \frac{6}{32} A \quad & | & A &=& 16\cdot 21 \\ &=& \frac{6}{32} \cdot 16\cdot 21 \\ &=& \frac{6}{2} \cdot 21 \\ &=& 3 \cdot 21 \\ \triangle CXY &=& 63 \\ \hline \end{array}\)
