Bunuel wrote:
If ∆ PQR and ∆ PRS are equilateral, what fraction of PQRS is shaded?
(A) 1/3
(B) 1/4
(C) 1/6
(D) 1/9
(E) 1/12
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Finding the answer to this question did not take a ton of time, but explaining that answer without a diagram is hard
1) SIX CONGRUENT SMALL TRIANGLES
Consider just ∆ PRS in figure PQRS
An equilateral triangle's altitude is also its median, angle bisector, and perpendicular bisector of base
All medians of an equilateral triangle are the same length:
Medians bisect the 60-degree angles to form two 30-degree angles
Medians bisect the bases into equal lengths and create right angles
Medians, therefore, create six identical 30-60-90 triangles
2) ANGLE MEASURES AND SIDE LENGTHS OF SIX TRIANGLES ARE CONGRUENT
Each triangle has:
60-degree angle at circumcenter, 90-degree angle at base bisector point, 30-degree angle at vertex
30-60-90 triangles have sides in ratio: \(x: x\sqrt{3}: 2x\)
See angles and side lengths in diagram of each small triangle. They are identical.
3) SHADED PORTION = \(\frac{1}{3}\)OF ONE TRIANGLE and \(\frac{1}{6}\) of figure PQRS
Because the six small triangles are equal,
one shaded portion equals \(\frac{1}{6}\) of area of ∆ PRS
There are two shaded portions. \(\frac{1}{6} + \frac{1}{6} = \frac{1}{3}\) of area of ∆ PRS
Given: ∆ PQR and ∆ PRS are equilateral
Length of PR = length of both PQ and QR
Areas, therefore, of ∆ PQR and ∆ PRS are equal
∆ PRS is \(\frac{1}{2}\) of the area of figure PQRS
Find shaded area's fraction of ∆ PRS and double the fraction
Shaded portion is therefore
\(\frac{1}{3} * \frac{1}{2} = \frac{1}{6}\)of the area of PQRS
Fraction of PQRS that is shaded = \(\frac{1}{6}\)
ANSWER C
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Any fool can know. The point is to understand. — Albert Einstein