adibapat
How is the answer for this A?
Shouldn't the triangle be equilateral as well to establish a proportion. Otherwise everything remains relative and dependant on the length of each side
Statement 1:
Attachment:
triangle BEF.png [ 59.2 KiB | Viewed 4593 times ]
In the figure above, E is the midpoint of BC and F is the midpoint of AB.
TRIANGLES ABC AND BEF:
Triangles ABC and BEF share angle EFB.
Shared angle EFB is formed by sides in the same ratio (BC:BE = 2:1, AB:BF = 2:1).
Rule:
Triangles with a shared angle (EBF) formed by corresponding sides in the same ratio (BC:BE = 2:1, AB:BF = 2:1) are SIMILAR.
Thus, triangle ABC is similar to triangle BEF.
RULE FOR TWO SIMILAR TRIANGLES:
If each side in the larger triangle is \(k\) times the corresponding side in the smaller triangle, then the area of the larger triangle is \(k^2\) times the area of the smaller triangle.
Here, since each side of ABC is 2 times the corresponding side in BEF, the area of ABC is 4 times the area of BEF.
Thus, \(BEF = \frac{1}{4}ABC\).
The same line of reasoning can be used to show that \(ADF = \frac{1}{4}ABC\) and that \(CDE = \frac{1}{4}ABC\), with the result that \(DEF = \frac{1}{4}ABC\).
Thus:
shaded region \(= DEF + BEF = \frac{1}{4}ABC + \frac{1}{4}ABC = \frac{1}{2}ABC = 50\)%
SUFFICIENT.