CRE 2 - Hexagon

We know internal angle of a regular polygon = $(n-2)\times \frac{180}{n}$

∴ Internal angle of a regular hexagon = $(6-2)\times \frac{180}{6}$ = 120°.

Hence, 120.

Since the given figure is symmetric, we have 12 smaller triangles all of whose area is same.

⇒ The hexagon is divided into 12 smaller triangles of equal areas.

∴ Area(Hexagon) = 12a

Also, Area(∆ACE) = 6a

⇒ Area(∆ACE) : Area(Hexagon) = 6a : 12a = 1 : 2.

Hence, option (c).

In the figure above, 

Area(∆AOB) = 1/6^th of area of hexagon = 6 cm².

Also, Area(∆AMB) = ½ of area ∆AOB = 3 cm².

Hence, (c).

Perimeter of a hexagon = 6 × side

∴ 6 × side = 36

⇒ Side of the hexagon = 6 cm.

Now, area of the regular hexagon = $6\times \left(\frac{\sqrt{3}}{4}\right){\mathrm{a}}^2$ = $\left(\frac{3\sqrt{3}}{2}\right)\times 6^2$ = 54√3.

Hence, option (c).

Suppose side of the hexagon = a.

In trapezium ABCF, CF = 2a and AB = a and R and Q are the mid points of oblique sides.

∴ RQ = (AB + CF)/2 = (a + 2a)/2 = 1.5a

⇒ $\frac{Area\left(∆PQR\right)}{Area\left(Hexagon\right)}$ = $\frac{\frac{\sqrt{3}}{4}{\left(1.5a\right)}^2}{6\times \frac{\sqrt{3}}{4}{a}^2}$ = $\frac{2.25}{6}=\frac{3}{8}$

Hence, option (a).

Let the side of the hexagon obtained be 'x'.

Since, DE || BC, ∆ADE will be an equilateral triangle.
∴ AD = DE = AE = x

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⇒ Side AB is divided equally in three parts, hence the side of the hexagon will be one-third that of the triangle.

∴ Side of the hexagon = 1/3 × 12 = 4 cm. 

⇒ Area of the hexagon = 6 × √3/4 × (4)² = 24√3

⇒ Perimeter of the hexagon = 6 × 4 = 24

Hence, option (b).