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26 Jan 2023, 02:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
54% (02:36) correct
46%
(02:05)
wrong
based on 46
sessions
History
Date
Time
Result
Not Attempted Yet
Uwe wants to split his rhombus-shaped garden into two triangular plots, one for planting strawberries and one for planting vegetables, by erecting a fence from one corner of the garden to the opposite corner. If one angle of the garden measures 60 degrees and one side of the garden measures x meters, which of the following could be the area of the vegetable plot?
I. \(\frac{x^2 \sqrt{3}}{2}\)
II. \(\frac{x^2 \sqrt{3}}{4}\)
III. \(\frac{x^2 \sqrt{3}}{8}\)
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II and III
I. \(\frac{x^2 \sqrt{3}}{2}\)
II. \(\frac{x^2 \sqrt{3}}{4}\)
III. \(\frac{x^2 \sqrt{3}}{8}\)
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II and III
26 Jan 2023, 22:35
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Bunuel
Given: One angle of the garden measures 60 degrees.
In a Rhombus the diagonals are perpendicular to each other.
We can have two possible cases as shown below -
Case 1
\(\angle ABC = 60\)
In \(\triangle BOA\)
- \(\angle BAO = 60\)
- BA = x
We have a 30 - 60 - 90 triangle
Hence
AO = OC = \(\frac{x}{2}\)
BO = vertical height = \(\frac{\sqrt{3}}{2} * x\)
Area of \(\triangle ABC\)
= \(\frac{1}{2}\) * AC * BO
= \(\frac{1}{2} * (\frac{x}{2} * 2) * \frac{\sqrt{3}}{2} \)
= \(\frac{\sqrt{3}}{4}x\)
Case 2
\(\angle QPS = 60\)
In \(\triangle QOP\)
- \(\angle PQO = 60\)
- PQ = x
We have a 30 - 60 - 90 triangle
Hence
QO = vertical height = \(\frac{x}{2}\)
PO = OR = \(\frac{\sqrt{3}}{2} * x\)
Area of \(\triangle PQR\)
= \(\frac{1}{2}\) * PR * QO
= \(\frac{1}{2} *(\frac{\sqrt{3}}{2} * 2) * \frac{x}{2} \)
= \(\frac{\sqrt{3}}{4}x\)
Option B
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15 Aug 2024, 19:12
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