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
According to the question,
Area of semi- circle = 77 m
(1/2) x ? x r2 = 77
? r2 = (77 x 2 x 7)/22
? r = 7m
?Circumference of semi- circle =?r + 2r
= ( ? + 2)r
= [(22/7) + 2] x 7
= 36 m
Distance covered in 1 revolution
= (44 x 1000) / 4000
= 11 m
According to the question,
2?r = 11
? (44/7) x r = 11
? r = (11 x 7) / 44 = 1.75 m
a1 = 68/4 = 17 cm
and a2 = 60/4 = 15 cm
[ where a1 and a2 are sides]
According to the question,
Area of the third square = [(17)2 - (15)2 ]
= (17 + 15) (17 - 15)
= 32 x 2
= 64 sq cm
Let a3 = Side of the third square.
According to the question, (a3)2 = 64 sq cm
? a3 = ?64 = 8 cm
? Perimeter of the third square = 4 x a3 = 4 x 8 = 32 cm.
According to the question,
4a = 68 [ where a = side]
?a = 68/4 = 17 cm
? Required area = a2
= (17)2 = 289 sq cm
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
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
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%
Let original radius be r.
Then, according to the questions,
? (r + 1)2 - ?r2 = 22
? ? x [(r + 1)2 - r2] = 22
? (22/7) x (r + 1 + r ) x (r + 1 - r) = 22
? 2r + 1 = 7
? 2r = 6
? r = 6/2 = 3 cm
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
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