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If AD is 9 and BD is 3
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19 Jan 2020, 02:43
19 Jan 2020, 02:43
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If AD is \(9\) and BD is \(3 \sqrt{3}\), what is the area of triangle ABC
A. \(\frac{27 \sqrt{3}}{2}\)
B. \(18 \sqrt{3}\)
C. \(27 \sqrt{3}\)
D. \(36 \sqrt{3}\)
E. It can not be determined
Kudos for the right answer and explanation
Re: If AD is 9 and BD is 3
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06 Feb 2020, 06:47
06 Feb 2020, 06:47
1
From ABD we can calculate that AB equals to \( \sqrt{81+27}\) =\( \sqrt{108}\) =\( 6\sqrt{3}\)
Let’s denote DC as x and BC as y then we get two equations:
(1) From ABC we get: y^2 = (9+x)^2- 108
(2) From BDC we get: y^2 = x^2 + 27
Simplifying equation (1) and substituting y^2 into it from equation (2) results in:
x^2 + 27 = 81 +18x +x^2 - 108
18x = 54
DC = x = 3
BC = y = \( \sqrt{9+27}\) = 6
The area of triangle ABC is then \( \frac{1}{2}*6*6\sqrt{3}\) = \( 18\sqrt{3}\)
Let’s denote DC as x and BC as y then we get two equations:
(1) From ABC we get: y^2 = (9+x)^2- 108
(2) From BDC we get: y^2 = x^2 + 27
Simplifying equation (1) and substituting y^2 into it from equation (2) results in:
x^2 + 27 = 81 +18x +x^2 - 108
18x = 54
DC = x = 3
BC = y = \( \sqrt{9+27}\) = 6
The area of triangle ABC is then \( \frac{1}{2}*6*6\sqrt{3}\) = \( 18\sqrt{3}\)
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If AD is 9 and BD is 3
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08 Feb 2021, 22:38
08 Feb 2021, 22:38
1
ADB is a 30-60-90 triangle:
Angle BAD = 30
Angle ABD = 60
Angle ADB = 90
Also, BDC is a 30-60-90 triangle:
Angle CBD = 30
Angle BCD = 60
Angle BDC = 90
In a 30-60-90 triangle, The ratio of 3 sides is x : \(\sqrt{3}\)x : 2x
i.e. opposite to 30 = x
opposite to 60 = \(\sqrt{3}\)x
opposite to 90 = 2x
Using the above property;
AB = 6\(\sqrt{3}\) and BC = 6
Therefore, Area of triangle ABC = \(\frac{1}{2}\) x 6\(\sqrt{3}\) x 6 = 18\(\sqrt{3}\)
Hence, option C
Angle BAD = 30
Angle ABD = 60
Angle ADB = 90
Also, BDC is a 30-60-90 triangle:
Angle CBD = 30
Angle BCD = 60
Angle BDC = 90
In a 30-60-90 triangle, The ratio of 3 sides is x : \(\sqrt{3}\)x : 2x
i.e. opposite to 30 = x
opposite to 60 = \(\sqrt{3}\)x
opposite to 90 = 2x
Using the above property;
AB = 6\(\sqrt{3}\) and BC = 6
Therefore, Area of triangle ABC = \(\frac{1}{2}\) x 6\(\sqrt{3}\) x 6 = 18\(\sqrt{3}\)
Hence, option C
