Question

A calf is tied with a rope of length 6 m at the corner of a square grassy lawn of side 20 m. If the length of the rope is increased by 5.5 m, find the increase in area of the grassy lawn in which the calf can graze.

Answer 1

Given: A calf is tied with a rope of length 6m at the corner of a square grassy lawn of side 20m.

So, radius of quadrant DPQD (r) = length of rope = 6 m

Therefore,

$AreaofsectorDPQD=\frac{\pi r^2\theta }{{360}^0}$

$=\frac{3.14\times (6{)}^2\times {90}^0}{{360}^0}$

$=0.785\times 36=28.26m^2$

Now, if the length of the rope is increased by 5.5 m

So, total length of the rope (R) = 6 + 5.5 = 11.5 m

$AreaofsectorDRSD=\frac{\pi R^2\theta }{{360}^0}$

$=\frac{3.14\times (11.5{)}^2\times {90}^0}{{360}^0}$

$=0.785\times 132.25=103.81625m^2$

Therefore,

Increased area = Area of sector DRSD – Area of sector DPQD

= 103.81625 – 28.26

= 75.55625 $m^2$

= 75.56 $m^2$

Hence, the increase in area of the grassy lawn is 75.56 $m^2$.