Question : The area of a regular hexagon with side '$a$' is:
Option 1: $\frac{3\sqrt3}{4}a^2$ sq. unit
Option 2: $\frac{12}{2\sqrt3}a^2$ sq. unit
Option 3: $\frac{9}{2\sqrt3}a^2$ sq. unit
Option 4: $\frac{6}{2\sqrt3}a^2$ sq. unit
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Correct Answer: $\frac{9}{2\sqrt3}a^2$ sq. unit
Solution : If we break a regular hexagon with side '$a$', then we will get 6 equilateral triangles with sides '$a$'. So, the area of a regular hexagon = 6 × (Area of an equilateral triangle) We know that, the area of an equilateral triangle with side '$a$' = $\frac{\sqrt3}{4}a^2$ Therefore, the area of the regular hexagon with side '$a$' = $6×\frac{\sqrt3}{4}a^2$ = $\frac{3\sqrt3}{2}a^2$ = $\frac{9}{2\sqrt3}a^2$ Hence, the correct answer is $\frac{9}{2\sqrt3}a^2$ sq. unit.
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Question : If $x=\sqrt3+\frac{1}{\sqrt3}$, then the value of $(x-\frac{\sqrt{126}}{\sqrt{42}})(x-\frac{1}{x-\frac{2\sqrt3}{3}})$ is:
Option 1: $\frac{5\sqrt3}{6}$
Option 2: $\frac{2\sqrt3}{3}$
Option 3: $\frac{5}{6}$
Option 4: $\frac{2}{3}$
Question : The base of a right pyramid is an equilateral triangle of side $10\sqrt3$ cm. If the total surface area of the pyramid is $270\sqrt3$ sq. cm. its height is:
Option 1: $12\sqrt3$ cm
Option 2: $10$ cm
Option 3: $10\sqrt3$ cm
Option 4: $12$ cm
Question : The value of $3\frac{1}{2} - [2\frac{1}{4}+ 1\frac{1}{4} - \frac{1}{2}(1\frac{1}{2}-\frac{1}{3} -\frac{1}{6})]$ is:
Option 1: $\frac{1}{2}$
Option 2: $2\frac{1}{2}$
Option 3: $3\frac{1}{2}$
Option 4: $9\frac{1}{2}$
Question : If the sides of an equilateral triangle are increased by 1 metre, then its area is increased by $\sqrt3$ sq. metre. The length of any of its sides is:
Option 1: $2$ metres
Option 2: $\frac{5}{2}$ metres
Option 3: $\frac{3}{2}$ metres
Option 4: $\sqrt3$ metres
Question : If $\sqrt{a}=3b$, then $\frac{a}{b^2}$ is equal to:
Option 1: $\frac{1}{6}$
Option 2: $6$
Option 3: $\frac{1}{9}$
Option 4: $9$
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