MATHS - CLASS X
AREAS RELATED TO CIRCLES
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5m long rope. Find:
(i) the area of that part of field which the horse can graze
(ii) the increase in the grazing are if the rope were 10 m long instead of 5m.
Answers
Answered by
17
hey mate
here is your answer
(I)-the area of the part of the field which the horse can graze=area of the quadrant whose radius is 5m
=>1/4*π*r²
=>1/4*π*5²
=>1/4*π*25 m²
in order to find the increase in the area ,we first need to find the area of the grazing field when the radio of the quadrant is 10m
=>1/4*π*r²
=>1/4*π*10²
=>1/4*π*100
=>25π
thus,the increase in the area of the grazing field=
=>25π-25π/4
=>3/4*25*π
=>18.75π
here is your answer
(I)-the area of the part of the field which the horse can graze=area of the quadrant whose radius is 5m
=>1/4*π*r²
=>1/4*π*5²
=>1/4*π*25 m²
in order to find the increase in the area ,we first need to find the area of the grazing field when the radio of the quadrant is 10m
=>1/4*π*r²
=>1/4*π*10²
=>1/4*π*100
=>25π
thus,the increase in the area of the grazing field=
=>25π-25π/4
=>3/4*25*π
=>18.75π
Answered by
57
Here's the answer
✔️ (Refer the first attachment for question 1)
✔️ (Refer the second attachment for question 2)
In Question 1 ,
See first attachment
Let ABCD be the square shaped field
Length of the rope is 5 cm
Let BG = r = 5 cm
Note that Angle GBH = 90° ( All angles of square are right angled )
So ,
According to the first question,
To find the area of that part of field which the horse can graze ,
We need to find area of sector GBH.
Area of sector GBH =![\sf { \frac{ \theta}{360} \times \pi \: r {}^{2}} \sf { \frac{ \theta}{360} \times \pi \: r {}^{2}}](https://tex.z-dn.net/?f=+%5Csf+%7B+%5Cfrac%7B+%5Ctheta%7D%7B360%7D+%5Ctimes+%5Cpi+%5C%3A+r+%7B%7D%5E%7B2%7D%7D)
![\sf{\therefore \ \frac{90}{360} \times 3.14 \times 5 {}^{2}} \sf{\therefore \ \frac{90}{360} \times 3.14 \times 5 {}^{2}}](https://tex.z-dn.net/?f=+%5Csf%7B%5Ctherefore+%5C+%5Cfrac%7B90%7D%7B360%7D+%5Ctimes+3.14+%5Ctimes+5+%7B%7D%5E%7B2%7D%7D+)
![\sf{ \frac{1}{4} \times 3.14 \times 5 \times 5} \sf{ \frac{1}{4} \times 3.14 \times 5 \times 5}](https://tex.z-dn.net/?f=+%5Csf%7B+%5Cfrac%7B1%7D%7B4%7D+%5Ctimes+3.14+%5Ctimes+5+%5Ctimes+5%7D)
![\sf{ = \frac{78.5}{4}} \sf{ = \frac{78.5}{4}}](https://tex.z-dn.net/?f=+%5Csf%7B+%3D+%5Cfrac%7B78.5%7D%7B4%7D%7D)
![\sf {= 19.625} \sf {= 19.625}](https://tex.z-dn.net/?f=+%5Csf+%7B%3D+19.625%7D)
Therefore ,
Area of the field where Horse can gaze is 19.625 sq. m.
Now,
Coming to your next question,
Question 2 -
=> Instead of 5 cm , rope is of 15 m
So , Rope's size was increased by 10 m
Now, Refer the second attachment
To find the area gazed by horse,
Find Area of sector TBS
Area of sector TBS![\sf{ = \frac{ \theta}{360} \times \pi \: r {}^{2}} \sf{ = \frac{ \theta}{360} \times \pi \: r {}^{2}}](https://tex.z-dn.net/?f=+%5Csf%7B+%3D+%5Cfrac%7B+%5Ctheta%7D%7B360%7D+%5Ctimes+%5Cpi+%5C%3A+r+%7B%7D%5E%7B2%7D%7D)
![\sf{ = \frac{90}{360} \times 3.14 \times 10 \times 10} \sf{ = \frac{90}{360} \times 3.14 \times 10 \times 10}](https://tex.z-dn.net/?f=+%5Csf%7B+%3D+%5Cfrac%7B90%7D%7B360%7D+%5Ctimes+3.14+%5Ctimes+10+%5Ctimes+10%7D)
![\sf{ = \frac{1}{4} \times 314} \sf{ = \frac{1}{4} \times 314}](https://tex.z-dn.net/?f=+%5Csf%7B+%3D+%5Cfrac%7B1%7D%7B4%7D+%5Ctimes+314%7D)
![\sf{= 78.5 \: m {}^{2}} \sf{= 78.5 \: m {}^{2}}](https://tex.z-dn.net/?f=+%5Csf%7B%3D+78.5+%5C%3A+m+%7B%7D%5E%7B2%7D%7D)
Therefore,
Area gazed by horse is 78.5 sq. m now.
Therefore,
Area gazed now - Area gazed earlier = Increase in the area for grazing
= Area of sector GBH - Area of TBS =
= 78.5 - 19.625
= 58.875 sq.m
Thus,
Increase in the area was 58.875 sq.m
Hence, Final answer =>
Question 1's answer
=> Area of the field where Horse can gaze is 19.625 sq. m.
Question 2's answer
=> Increase in the area was 58.875 sq.m
_____________________________
Thanks!!
If any doubt , write it in the comments box :)
Hope it helps :)
___________________________
✔️ (Refer the first attachment for question 1)
✔️ (Refer the second attachment for question 2)
In Question 1 ,
See first attachment
Let ABCD be the square shaped field
Length of the rope is 5 cm
Let BG = r = 5 cm
Note that Angle GBH = 90° ( All angles of square are right angled )
So ,
According to the first question,
To find the area of that part of field which the horse can graze ,
We need to find area of sector GBH.
Area of sector GBH =
Therefore ,
Area of the field where Horse can gaze is 19.625 sq. m.
Now,
Coming to your next question,
Question 2 -
=> Instead of 5 cm , rope is of 15 m
So , Rope's size was increased by 10 m
Now, Refer the second attachment
To find the area gazed by horse,
Find Area of sector TBS
Area of sector TBS
Therefore,
Area gazed by horse is 78.5 sq. m now.
Therefore,
Area gazed now - Area gazed earlier = Increase in the area for grazing
= Area of sector GBH - Area of TBS =
= 78.5 - 19.625
= 58.875 sq.m
Thus,
Increase in the area was 58.875 sq.m
Hence, Final answer =>
Question 1's answer
=> Area of the field where Horse can gaze is 19.625 sq. m.
Question 2's answer
=> Increase in the area was 58.875 sq.m
_____________________________
Thanks!!
If any doubt , write it in the comments box :)
Hope it helps :)
___________________________
Attachments:
![](https://hi-static.z-dn.net/files/dd7/6c3d2825417697df6fe08fb902d70118.jpg)
![](https://hi-static.z-dn.net/files/db3/425b827ac66f7375b7117df31ea1bed4.jpg)
Anonymous:
awesome answer bro
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