Let circumference = 100 cm .
Then, ? 2?r = 100
? r = 100/2?
=50/?
? New circumference
= 105 cm
Then, 2?R = 105
? R = 105 / (2?)
&rArr Original area = [ ? x (50/?) x (50/?) ]
= 2500/? cm2
? New Area = [? x (105/2?) x (105/2?)]
= 11025 / (4?) cm2
? Increase in area = [11025/(4?)] - 2500/? cm2
= 1025 / 4? cm2
Required increase percent [1025/(4?)] x 2500/? x 100 = 41/4%
= 10.25%
? 1/2 x 2 ? x h
= ?3/4 x (2 ?3)2
? h = 3 cm.
Circumference of the circle = 2?r = 25
? r = 25 / (2?)
According to the formula,
Side of inscribed square = r?2
= 25 / (2?r x ?2)
= 25 / (??2 ) cm
Area of square = 64 sq cm
(Side)2 = 64
? Side = ?64 = 8 cm
According to the question,
? 2?r = 4 x 8
? r = (4 x 8)/2?
? r = 16/?
? Area of the circle
= ? x (16/?) x (16/?) sq cm
= 256/? sq cm.
Diameter of the circle = 13 + 5 = 18 cm
? Radius = Diameter/2 =18/2 = 9 cm
Area of the circle = ?r2 = (22/7) x 92
= (22 x 81)/9
= 1782/7
= 254.57 sq cm
= 255 sq cm
Let the sides of trapezium be 5k and 3k, respectively
According to the question,
(1/2) x [(5k + 3k) x 12] = 384
? 8k = (384 x 2)/12 = 64
? k = 64/8 = 8 cm
Length of smaller of the parallel sides = 8 x 3 = 24 cm
Angle swept in 30 min= 180°
Area swept = [(22/7) x 7 x 7] x [180°/360°] cm2
= 77 cm2
? 22/7 x r2 = 462
? r2 = (462 x 7) /22 = 147
? r = 7?3 cm
? Height of the triangle = 3r = 21?3 cm
Now, ? a2 = a2/4 + (3r)2
? 3a2/4 = (21?3)2
? a2 = (1323 x 4)/3
? a = 21x 2 = 42 cm
? Perimeter = 3a = 3 x 42
=126 cm
Area of park = 100 x 100 = 10000 m2
Area of circular lawn = Area of park - area of park excluding circular lawn
= 10000 - 8614
= 1386
Now again area of circular lawn = (22/7) x r2 = 1386 m2
? r2 = (1386 x 7) / 22
= 63 x 7
= 3 x 3 x 7 x 7
? r = 21 m
Area of the room =(544 x 374) cm2
size of largest square tile = H.C.F. of 544 & 374
= 34 cm
Area of 1 tile = (34 x 34) cm2
? Least number of tiles required
= (544 x 374) / (34 x 34) = 176
Ratio of similar triangle
= Ratio of the square of corresponding sides
= (3x)2 / (4x)2 = 9x2 / 16x2
= 9/16 = 9 : 16
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