Ozlemg wrote:A triangle ABC, point X is the midpoint of side AC and point Y is the midpoint of side BC. If point R is the midpoint of line segment XC and if point S is the midpoint of line segment YC, what is the area of triangle region RCS
(1) the area of triangular region ABX is 32
(2) the length of one of the altitudes of triangle ABC is 8
Draw the figure:
Triangle RCS is similar to triangle ABC for the following reasons:
RC = (1/4)AC
CS = (1/4)BC
The two triangles share angle BCA.
Since corresponding sides yield the same ratio, and the angle between these corresponding sides is shared by the two triangles, the two triangles are similar.
Area of triangle ABC = (1/2)bh.
Since each corresponding side of RCS is 1/4 the corresponding side of ABC, the base of RCS is 1/4 the base of ABC, and the height of RCS is 1/4 the height of ABC.
Thus, the area of RCS = (1/2)*(1/4)b*(1/4)h = (1/16)(1/2)bh, indicating that the area of RCS is 1/16 the area of ABC.
Question rephrased: What is the area of ABC?
Statement 1: ABX = 32.
When a line is drawn from a vertex to the midpoint of the opposite side, the triangle is split into two equal areas.
Notice that AX = b/2.
Since the area of ABC = (1/2)bh, we know that the area of ABX = (1/2)(b/2)h, yielding half the area of ABC.
Since ABX = 32, we know that ABC = 64 and that RCS = (1/16)(64) = 4.
Sufficient.
Statement 2: h = 8.
No way to determine the area of ABC.
Insufficient.
The correct answer is
A.
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