A cow is tied to a corner (vertex) of a regul | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The interior angle of a regular hexagon is $120^\circ$. The exterior angle is $360^\circ - 120^\circ = 240^\circ$. The cow can graze in a sector o

Vedclass · Accepted Answer

$\frac{5}{2} \pi a^2$

Vedclass · Answer

(B) The interior angle of a regular hexagon is $120^\circ$. The exterior angle is $360^\circ - 120^\circ = 240^\circ$. The cow can graze in a sector of radius $R = \frac{5a}{2}$ and angle $240^\circ$. As the rope wraps around the vertices of the hexagon,the radius decreases by $a$ at each turn. $1$. Sector $1$: Radius $R_1 = \frac{5a}{2}$,Angle $	heta_1 = 240^\circ = \frac{2\pi}{3}$ radians. Area $A_1 = \frac{1}{2} R_1^2 	heta_1 = \frac{1}{2} (\frac{5a}{2})^2 (\frac{2\pi}{3}) = \frac{25\pi a^2}{12}$. $2$. Sector $2$: Radius $R_2 = \frac{5a}{2} - a = \frac{3a}{2}$,Angle $	heta_2 = 60^\circ = \frac{\pi}{3}$ radians. Area $A_2 = \frac{1}{2} R_2^2 	heta_2 = \frac{1}{2} (\frac{3a}{2})^2 (\frac{\pi}{3}) = \frac{3\pi a^2}{8}$. $3$. Sector $3$: Radius $R_3 = \frac{3a}{2} - a = \frac{a}{2}$,Angle $	heta_3 = 60^\circ = \frac{\pi}{3}$ radians. Area $A_3 = \frac{1}{2} R_3^2 	heta_3 = \frac{1}{2} (\frac{a}{2})^2 (\frac{\pi}{3}) = \frac{\pi a^2}{24}$. Total Area $= A_1 + A_2 + A_3 = \pi a^2 (\frac{25}{12} + \frac{3}{8} + \frac{1}{24}) = \pi a^2 (\frac{50 + 9 + 1}{24}) = \pi a^2 (\frac{60}{24}) = \frac{5}{2} \pi a^2$.

$A$ cow is tied to a corner (vertex) of a regular hexagonal fenced area of side $a \ m$ by a rope of length $\frac{5a}{2} \ m$ in a grass field. (The cow cannot graze inside the fenced area). What is the maximum possible area of the grass field to which the cow has access to graze?

Similar Questions

If the distance of a variable point $P(x, y)$ from a point $A(2, -2)$ is twice the distance of $P$ from the $Y$-axis,then the equation of the locus of $P$ is:

If a point $P(x, y)$ moves such that the sum of the squares of its coordinates is equal to their product,then the locus of $P$ excluding the origin is

The locus of the centre of circles passing through $(a, b)$ and cutting the circle $x^2+y^2-2x+4y-4=0$ orthogonally is

The locus of the centre of a variable circle which cuts the circles $x^2 + y^2 - 2x - 4y - 1 = 0$ and $x^2 + y^2 - 4x - 2y - 1 = 0$ orthogonally is:

If the intercepts made by a variable circle on the $X$-axis and $Y$-axis are $8$ and $6$ units respectively,then the locus of the centre of the circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module