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14 Feb 2022, 05:15
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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:
75% (01:47) correct
25%
(01:44)
wrong
based on 64
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Triangle ABC is an equilateral triangle with an altitude of 6. What is its area?
A. \(4\sqrt{3}\)
B. \(6\sqrt{3}\)
C. \(12\sqrt{3}\)
D. \(24\sqrt{3}\)
E. \(36\sqrt{3}\)
Attachment:
gmhtdbasr_img3.png [ 2.62 KiB | Viewed 2167 times ]
kungfury42
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14 Feb 2022, 21:22
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AB*cos(ABD) = AD
Since ABC is an equilateral triangle (angle A = angle B = angle C = 60°), we know that the altitude cuts it in two symmetrical halves, therefore, angle ABD = angle B ÷ 2 = 60/2 = 30°
AB*cos(30°) = AD
AB*(√3/2) = 6
AB = 12/√3 = 4√3
Since it is an equilateral triangle, AB=BC=AC = 4√3
Now, area of triangle ABC = ½*Base*Height = ½*AC*BD = ½*4√3*6 = 12√3
Therefore, option C is our answer.
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Since ABC is an equilateral triangle (angle A = angle B = angle C = 60°), we know that the altitude cuts it in two symmetrical halves, therefore, angle ABD = angle B ÷ 2 = 60/2 = 30°
AB*cos(30°) = AD
AB*(√3/2) = 6
AB = 12/√3 = 4√3
Since it is an equilateral triangle, AB=BC=AC = 4√3
Now, area of triangle ABC = ½*Base*Height = ½*AC*BD = ½*4√3*6 = 12√3
Therefore, option C is our answer.
Posted from my mobile device
