Math, asked by Anny121, 1 year ago

Hey Users!!

^Answer this Question^

●Content Quality Required●

◆No Spamming◆

Attachments:

Answers

Answered by Anonymous
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
Answered by HarishAS
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.

Diagonal \ square =  side\sqrt{2}

 = 56  \sqrt{2}\ m^{2} 

We know that ,

OA=OB=OC=OD=  \frac{1}{2}  \times Diagonal = \frac{1}{2} \times 56 \sqrt{2} =28\sqrt{2}

Now,

i) Ar\ of\ AOD=Ar\ of\ BOC =  \frac{1}{2}  \times base\times height =  \frac{1}{2}\times 28 \sqrt{2}\times 28 \sqrt{2}

→ 28^{2} = 784m^{2}

ii) Ar\ of\ sector\ AOB = Ar\ of\ sector\ COD =   \frac{\theta}{360} \times  \pi \times r^{2}

Here, 

r = Length of OB or Length of OC . 

and  θ = ∠AOB = 90°

So, 

Area\ of \ sector =  \frac{90}{360} \times \frac{22}{7} \times (28\sqrt{2})^{2}

→ =  \frac{90}{360} \times  \frac{22}{7}  \times 1568

→ = 1 \times 22 \times 56

= 1232 \ m^{2}

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.
Anny121: it is not copy [email protected] has written it himself
HarishAS: Really i promise.
Anny121: i know bro..
HarishAS: Ok ....
Anny121: u have written it urself...☺️
Anny121: ur answer and harish's answer is completely different....u both have written urselves
Anny121: ___no more comments__
Similar questions