GeminiHeat wrote:
In the diagram, 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:
The attachment isofindx.jpg is no longer available
Not that easy as it appears to be!
GeminiHeat Good question
Drop a perpendicular CD on AB (Refer to the figure below)
Which will give us ACD as 30-60-90 Right triangle
i.e. opposite to 30 we will have side \(a\) (let), opposite to 60 we will have side \(\sqrt{3}a\), and opposite to 90 we will have side \(2a\)
We already know, \(2a = x\), So \(a = \frac{x}{2}\)
Therefore, CD = \(\frac{x}{2}\) and AD = \(\frac{\sqrt{3}x}{2}\)
Now, AB = AD + DB
\(x = \frac{\sqrt{3}x}{2}\) + DB
DB = \(x - \frac{\sqrt{3}x}{2}\)
DB = \(\frac{(2 - \sqrt{3})x}{2}\)
In triangle, CDB;
\(BC^2 = DB^2 + CD^2\)
\((\sqrt{2})^2 = [\frac{(2 - \sqrt{3})x}{2}]^2 + (\frac{x}{2})^2\)
\(2 = \frac{x^2}{4} + \frac{(4 + 3 - 4\sqrt{3})x^2}{4}\)
\(2 = \frac{x^2}{4} + \frac{7x^2}{4} - \sqrt{3}x^2\)
\(2 = 2x^2 - \sqrt{3}x^2 = (2 - \sqrt{3})x^2\)
\(x^2 = \frac{2}{(2 - \sqrt{3})}\)
Upon rationalising;
\(x^2 = \frac{2(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}\)
\(x^2 = 4 + 2\sqrt{3}\)
You can either use the calculator or square the answer choices! You will get \(x = 1+\sqrt{3}\)
Hence, option B
Attachments
In the diagram, what is the value of x.png [ 21.69 KiB | Viewed 1946 times ]