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22 Sep 2017, 00:32
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
73% (01:41) correct
27%
(02:04)
wrong
based on 142
sessions
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Not Attempted Yet
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
Attachment:
2017-09-20_1023.png [ 10.15 KiB | Viewed 8474 times ]
22 Sep 2017, 03:19
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Bunuel
Hi...
∆PQR and ∆ PRS are equilateral and same as they share common side PR..
Concentrate now on ∆PRS..
The altitude from the vertices intersect at 1/3 of altitude/height.
So the unshaded triangles along PR gave 1/3 the area of ∆PRS.
The remaining portion is 2/3 of area.
AND it can be seen that the shaded portion is 1/2 of this so 1/2 * 2/3 =1/3 of ∆PRS..
But quad PQRS = PRS + PQS=2 times area of PRS
So shaded portion is 1/3 * 1/2= 1/6 of PQRS
C
22 Sep 2017, 10:14
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Bunuel
Attachment:
equitriangleshaded.png [ 13.32 KiB | Viewed 7415 times ]
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
