guyz plz solve 3rd 4th 5th 6th sums
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3) Let the radius of the circle is r
Now diameter of the circle = 2r
The circumference of the circle = 2πr
Now according to question,
2πr = 2r + 75
=> 2πr - 2r = 75
=> 2r(π - 1) = 75
=> 2r(22/7 - 1) = 75
=> 2r(22 - 7)/7 = 75
=> 2r(22 - 7) = 75*7
=> 2r*15 = 75*7
=> 2r = (75*7)/15
=> 2r = 5*7
=> 2r = 35
=> r = 35/2
So, radius of the circle = 35/2 = 17.5 cm
and diameter of the circle = 2r = 2* 35/2 = 35 cm
Now diameter of the circle = 2r
The circumference of the circle = 2πr
Now according to question,
2πr = 2r + 75
=> 2πr - 2r = 75
=> 2r(π - 1) = 75
=> 2r(22/7 - 1) = 75
=> 2r(22 - 7)/7 = 75
=> 2r(22 - 7) = 75*7
=> 2r*15 = 75*7
=> 2r = (75*7)/15
=> 2r = 5*7
=> 2r = 35
=> r = 35/2
So, radius of the circle = 35/2 = 17.5 cm
and diameter of the circle = 2r = 2* 35/2 = 35 cm
srini8381:
Hope it helps you
Answered by
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3. ATQ, the circumference of a circle exceeds it's diameter by 75cm.
let the radius of the circle be 'r'
we know that diameter is double of radius. therefore it's diameter is '2r'
now, it's given the circumference is 75cm more than it's diameter.
circumference of a circle = 2πr
=> 75 + 2r = 2πr
=> 75 = 2πr - 2r
=> 75 = r(2 × 22/7 - 2)
=> 75 = r(44/7 - 2)
=> 75 = r(44/7 - 14/7)
=> 75 = r(30/7)
=> 75/1 × 7/30 = r
=> 2.5 × 7 = r
=> r = 17.5cm
hence, the radius of the circle is 17.5cm
4. given radius of the circular field = 56m
we have to find the length of wire required to fence it two times.
so first of all, let's find the circumference of the circular field..
circumference of the circular field = 2πr
= 2 × 22/7 × 56
= 44 × 8
= 352m
since we have to fence it twice. multiply the circumference by 2.
hence, total length required to fence twice = 352 × 2 = 704m
5. given diameter :- 84m
therefore circumference of the circular garden = πd (u can also use 2πr by dividing diameter by 2 and u will get the radius)
= 22/7 × 84
= 22 × 12
= 264m
therefore total cost of the wire required @ rupees 10.85 = 264 × 10.85
= rupees 2864.4
6. circumference of inner track = 616m
therefore 2πr = 616m
=> 2 × 22/7 × r = 616m
=> 44/7 × r = 616m
=> r = 616/1 × 7/44
=> r = 14 × 7
=> r = 98
circumference of the outer track = 660m
=> 2πr = 660m
=> 2 × 22/7 × r = 660m
=> 44/7 × r = 660m
=> r = 660/1 × 7/44
=> r = 15 × 7
=> r = 105m
hence, width of the track = 105m - 98m
= 7m
HOPE THIS HELPS!!
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