Questions & Answers

Question

Answers

(i)the area of that part of that part of the field in which the horse can graze

(ii)The increase in the grazing area if the rope were 10 m long instead of 5m (Use =3.14)

Answer

Verified

128.7k+ views

In square all angles are \[90\] Hence\[\angle QBP = 90\]

In II nd part length of rope is increased to \[10m\]

Area gazed by horse now \[ = \] Area of sector HBG

And increasing in grazing Area\[ = \]Area grazed by horse Now-Area grazed previously

Let ABCD be square filed and the length of rope \[ = 5m(r)\] here the area of the field which a horse can graze is nothing but the Area of sector QBP.

(i)Here we have to find the part of that part of the field in which the horse can graze

Area of sector \[ = \dfrac{\theta }{{360}} \times \pi {r^2}\]

We can call it sector in which horse graze the area

Where r=length of rope

\[\theta = {90^0},\pi = \dfrac{{22}}{7}\]

On substituting the value in \[eq(i)\] we get

Area of sector \[ = \dfrac{{90}}{{360}} \times \dfrac{{22}}{7} \times {5^2}\]

\[ = 19.625{m^2}\]

(ii)If the rope is increased by the \[10m\] long the the area of the area gazed by horse now \[ = \] Area of sector HBG

So, new area \[ = \dfrac{{90}}{{360}} \times \dfrac{{22}}{7} \times {(10)^2}\]

\[ = 78.5{m^2}\]

We have to find increased in area which is nothing but subtracting the old area from new area

Increased area \[ = \]New area-old area

Increased area\[ = 78.5 - 19.625\]

Increased area\[ = 58.875{m^2}\]

The axes of a two-dimensional Cartesian system divide the plane into four infinite regions, called quadrants, each bounded by the two half-axes. Quadrant consists of four coordinates.

Area of the Quadrant:

\[ = \dfrac{{\pi \times {r^2}}}{4}\]

\[ = \dfrac{{3.14 \times 5 \times 5}}{4}\]

\[ = 19.625{m^2}\]------->(old area)

If length of the rope is increased

Area\[ = 78.5{m^2}\]------>(new area)

Therefore to get increased area we will subtract old area from new area i.e

Increased area \[ = \]New area-old area

Increased area\[ = 78.5 - 19.625\]

Increased area\[ = 58.875{m^2}\]