Bunuel wrote:
In right triangle \(\triangle ABC\), \(∠C\) is a right angle; \(\overline{DE}\) is parallel to \(\overline {AB}\); and D and E are the midpoints of \(\overline {CA}\) and \(\overline{ CB}\) , respectively. Each shaded triangle is similar to \(\triangle ABC\), and the shaded triangles are congruent to one another. What fraction of the area of \(\triangle ABC\) is shaded?
A. 1/8
B. 1/4
C. 1/3
D. 1/2
E. 3/4
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When mid points of two sides are joined, the line is parallel to the third side and the smaller triangle obtained is similar to the large triangle.
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Screenshot 2023-04-03 at 10.44.56 AM.png [ 20.76 KiB | Viewed 1150 times ]
So when D and E are joined, triangle CDE is similar to triangle CAB (sides and altitude ratios 1 : 2) so the ratio of their areas is 1 : 4.
Area of triangle CDE = (1/2) * CT * DE = x ....... (I)
So area of triangle ABC = 4x
Area of rectangle DEMP = DE * TN (DP and EM are perpendicular to the base because the shaded triangles are similar to triangle ABC)
Since TN = CT (because altitude ratios are 1 : 2 so CT : CN = 1:2)
Area of rectangle DEMP = DE * CT = 2x (from (I) above)
Since the shaded triangles are congruent, K is the mid point of PM and hence, area of shaded triangles is half the area of rectangle DEMP i.e. area of shaded triangles = x
Shaded area/ Area of ABC = x/4x = 1/4
Answer (B)