in Quantitative Aptitude retagged by
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Suppose the length of each side of a regular hexagon $\text{ABCDEF}$ is $2 \; \text{cm}.$ It $\text{T}$ is the mid point of $\text{CD},$ then the length of $\text{AT, in cm},$ is

  1. $\sqrt{15}$
  2. $\sqrt{13}$
  3. $\sqrt{12}$
  4. $\sqrt{14}$
in Quantitative Aptitude retagged by
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1 Answer

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We can draw the regular hexagon.

Every angle in a regular hexagon will be equal to $120^{\circ}$

Let the length of $\text{AC}$ be $x$ cm.

In $\triangle \text{ABC}$,

$x^{2}=2^{2} + 2^{2} – 2 \times 2 \times 2 \times \cos 120^{\circ}$

$\Rightarrow x^{2} = 4+4 – 8(\frac{-1}{2})$

$\Rightarrow x^{2} = 8+4$

$\Rightarrow x^{2} = 12$

$\Rightarrow \boxed{x = \sqrt{12}\;\text{cm}}$

The $\triangle\text{ACT}$ is right-angle triangle.

 We can apply the Pythagorean theorem.

 $\text{(Hypotenuse)}^{2} = \text{(Perpendicular)}^{2} + \text{(Base)}^{2}$

$\Rightarrow \text{(AT)}^{2} = \text{(AC)}^{2} + \text{(CT)}^{2}$

$\Rightarrow \text{(AT)}^{2} = (\sqrt{12})^{2} + (1)^{2}$

$\Rightarrow \text{(AT)}^{2} = 12+1$

$\Rightarrow \text{(AT)}^{2} = 13$

$\Rightarrow \boxed{\text{AT} = \sqrt{13} \; \text{cm}}$

Correct Answer $:\text{B}$

$\textbf{PS:}\;\text{Cosine Rule (Law of Cosines):}$ Given the following triangle $\text{ABC}$ with corresponding sides length $a, b,$ and $c:$

the law of cosines states that

$$\begin{aligned} a^2&=b^2+c^2-2bc \cdot \cos A\\ b^2&=a^2+c^2-2ac \cdot \cos B\\ c^2&=a^2+b^2-2ab \cdot \cos C. \end{aligned}​$$

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