The area of a circle is increased by 22 sq cm when its radius is increased by 1 cm. Find the original radius of the circle.

Increase in circumference of circle = 5%
? Increase in radius is also 5%.
Now, increase in area of circle = 2a + (a²/100) %
Where, a = increase in radius= 2 x 5 + (5 x 5)/100 % = 10.25%

Ratio of the areas of the circumcircle and incircle of a square
= [(Diagonal)²?] / [(Side)²?]
= [(Side x ?2)²] / (Side)² = 2/1 or 2 : 1

Distance covered in 1 round
= (Total Distance) / (Total round)
= (1.76 x 1000)/350
= 176/35 m

? 2?r = (1.76 x 100) / 35 cm
2? = Diameter = 17600 x 7/22 x 35 = 160 cm

Given that, a = 6
According to the formula,
Percentage increase in area
= 2a + [a²/100]%
= 2 x 6 + [36/100]%
= (12 + 0.36)%
= 12.36%

Let the radius of the park be r, then
?r + 2r = 288
(? + 2)r = 288
? [(22/7) + 2]r = 288
? r = (288 x 7)/36 = 56

? Area of the park = (1/2)?r²
= (1/2) x (22/7) x 56 x 56
= 4928

Length of to the rope = Radius of circle
According to the question,
?r² = 154
? r² = 154 x (7/22) = 7 x 7 = 49
? r = ?49 = 7 m

The man takes 3600 s for 14.4 km
The man will take 88 s for
14.4 x (88/3600) = 352/1000 km = 352 m

Now, circumference of circular field = 352 m
? 2?r = 352 m
2 x (22/7) x r = 352
? r = 56 m

Therefore, area of the field = ?r²
= (22/7) x 56 x 56
= 8 x 22 x 56 m²
= 9856 sq m.

Let the radius of circular field = r m.
Speed of person in m/s = 30/60 = 1/2m/s
According to the question,
[(2?r) /(1/2)] - [(2r)/(1/2)] = 30
? 4?r - 4r = 30
? [4 x (22/7) - 4]r =30
? (125 - 4)r = 30
? (8.5)r = 30
? r = 30/8.5 = 3.5 m

Length ot rectangle = 40 m
Let breadth of = k
Then, according of the question,
(40 + k)15 = 40k
? 600 + 15k = 40k
? 25k = 600
? k = 24 m

Let the length of rectangle = L m
? Breadth of rectangle = B m
Using conditions from the question,
L - B = 23 ....(i)
2(L + B) = 206
L + B = 103 ....(ii)

On adding Eqs. (i) and (ii), we get
2L = 126
? L = 63 m
? B = 103 - 63 = 40 m

Then , area of rectangle = L x B
= 63 x 40
= 2520 m² .