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In the diagram to the right, what is the value of x?
A. \(1+\sqrt 2\)
B. \(1+\sqrt 3\)
C. \(2\sqrt 2\)
D. \(\sqrt 2 + \sqrt 3\)
E. \(2\sqrt 3\)
Attachment:
Untitled.png [ 28.01 KiB | Viewed 15209 times ]
31 Mar 2012, 14:16
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In the diagram to the right, what is the value of x?
A. \(1+\sqrt 2\)
B. \(1+\sqrt 3\)
C. \(2\sqrt 2\)
D. \(\sqrt 2 + \sqrt 3\)
E. \(2\sqrt 3\)
Check below:
Triangle BDC is a 30°-60°-90° right triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\), (the leg opposite 30° (BD) corresponds with \(1\), the leg opposite 60° (DC) corresponds with \(\sqrt{3}\), and the hypotenuse \(BC=x\) corresponds with 2). Thus, since \(BC=x\) then \(BD=\frac{x}{2}\) and \(DC=\frac{x\sqrt{3}}{2}\).
Also notice that \(AD=AC-DC=x-\frac{x\sqrt{3}}{2}\).
Next, from Pythagoras theorem \(AB^2=BD^2+AD^2\);
\(2=(\frac{x}{2})^2+(x-\frac{x\sqrt{3}}{2})^2\);
\(2=x^2(\frac{1}{4}+1-2*\frac{\sqrt{3}}{2}+\frac{3}{4})\);
\(2=x^2(2-\sqrt{3})\);
\(x^2=\frac{2}{2-\sqrt{3}}\);
Rationalize: \(x^2=\frac{2*(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}=4+2\sqrt{3}\);
\(x^2=4+2\sqrt{3}=(\sqrt{3})^2+2\sqrt{3}+1=(\sqrt{3}+1)^2\);
\(x=\sqrt{3}+1\).
Answer: B.
Hope it's clear.
Attachment:
Ttriangle.png [ 23.26 KiB | Viewed 31524 times ]
06 Jun 2011, 19:28
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In the diagram to the right, what is the value of x?
A. \(1+\sqrt 2\)
B. \(1+\sqrt 3\)
C. \(2\sqrt 2\)
D. \(\sqrt 2 + \sqrt 3\)
E. \(2\sqrt 3\)
They have given a triangle in the question and a 30 degree angle but no right triangle. So, my first instinct is to drop an altitude (a) to get a 30-60-90 triangle.
Attachment:
Ques2.jpg [ 7.13 KiB | Viewed 14664 times ]
We know that the sides of a 30-60-90 triangle are \(a\), \(\sqrt{3}a\) and \(2a\).
Hence the smaller right triangle has sides \(a\), \(2a - \sqrt{3}a\) and \(\sqrt{2}\).
Using Pythagorean theorem,
\(a^2 + (2a - \sqrt{3}a)^2 = \sqrt{2}^2\).
\(a^2 = \frac{1}{4 - 2\sqrt{3}} = \frac{1}{(\sqrt{3} - 1)^2}\)
\(a = x/2 = \frac{1}{(\sqrt{3} - 1)}\)
\(x = \sqrt{3} + 1\)
