Hey Users!!
^Answer this Question^
●Content Quality Required●
◆No Spamming◆
Attachments:
![](https://hi-static.z-dn.net/files/d6f/b646454e9d9b8f4b19a6c29b28d56a6c.jpg)
Answers
Answered by
4
area of lawn : 56^2=3136cm^2
area of flower bed =180/360×22/7 × 56/2×56/2
=22×56=1232
so so flower bed 2×1232=2464
so lawn + flower bed=2464+3136=5600cm^2
area of arc = theta/360 ×πr^2
area of flower bed =180/360×22/7 × 56/2×56/2
=22×56=1232
so so flower bed 2×1232=2464
so lawn + flower bed=2464+3136=5600cm^2
area of arc = theta/360 ×πr^2
Answered by
14
Hey friend, Harish here.
Here is your answer:
Given that,
1) ABCD is a square with sides length 56m.
2) O is the centre of the arcs AB & CD.
To find,
The area of whole flower bed and lawn.
Solution,
Area of lawn and circular bed = Area of sector AOB + Area of sector DOC + Area of ΔCOB + Area of ΔAOD.
→
We know that ,
Now,
i)
→
ii)
Here,
r = Length of OB or Length of OC .
and θ = ∠AOB = 90°
So,
→
→
→
Then,
Total area = Area of sector AOB + Area of sector DOC + Area of ΔCOB + Area of ΔAOD.
= 1232 + 1232 + 784 +784
= 4032 m^{2}
Therefore Total area of lawn and flower bed is 4032m².
__________________________________________________
Hope my answer helps you.
HarishAS:
Who said, I was typin it.
[email protected] has written it himself
it is not copy Similar questions