Draw a Circle With Shaded Region
FIND Expanse OF THE SHADED REGION OF A Circumvolve
Example 1 :
The inner part of an athletics runway is backyard. Find the area of the lawn.
Solution :
Area of backyard
= Area of rectangle + 2(Area of semicircle)
Length = 100 m, width = 65 m
radius = 65/2 ==> 32.5 chiliad
Area of rectangular lawn = Length × width + 2(∏r 2 /ii)
= (100 × 65) + [22/vii × 32.5 × 32.5]
= 6500 + 3319.64
= 9819.6 m 2
Expanse of shaded region = 9820 m2
So, area of lawn is 9820 chiliad2.
Case two :
A door has the dimensions shown.
(a) How loftier is the door at its highest point ?
(b) What is the area of the door in square meters ?
Solution :
(a) From the picture, the door consist of rectangle and a semicircle.
= (180 + 40) cm
= 220 cm
220 cm high is the door at its highest bespeak.
(b)
Area of shaded region
= Area of rectangular door + Area of semicircle
Length = 180 cm ==> ane.eight 1000
Width = 80 cm ==> 0.8 chiliad
Diameter (d) = 80 cm
Radius = 40 cm ==> 0.40 m
= (1.8 × 0.8) + [(22/7 × 0.4 × 0.four)/2]
= 1.44 + 0.2514
= 1.69 m2
So, area of the door is 1.69 mii.
Example 3 :
Discover the expanse of the shaded region.
Solution :
Area of shaded region
= Surface area of rectangle - Area of circle
Length = four m, Width = 2.6 m
d = 2 m and r = 1 m
Area of shaded region = Length × width - ∏r two
= (4 × 2.vi) - ( 22/7 × 1 × i)
= (10.4 – 3.14) m 2
Expanse of shaded region = 7.26 one thousand 2
And so, area of the shaded region given above is7.26 1000 2.
Example 4 :
A circular table summit has a bore of 1.6 m. A rectangular tablecloth 2 g b y 2 m is placed over the table superlative. What area of the tablecloth overlaps the table ?
Solution :
Overlapping area
= Area of rectangular tablecloth - Area of circular tabular array peak
Length = 2 m, width = 2 1000
Diameter d = 1.6 k and radius = 0.viii one thousand
Overlapping area = Length × width - ∏r 2
= (2 ten 2) -(22/7 × 0.eight × 0.8)
= (4 - 2.0114) mtwo
Overlapping area = 1.98 gii
So, Area of the tablecloth is 1.98 mii
Example v :
Consider a square within a circle.
(a) Detect the area of:
i) the circle
ii) the shaded triangle
iii) the square.
(b) What percentage of the circle is occupied by the square ?
Solution :
(i) From the picture given radius of the circle is x cm.
Area of circle = ∏r2
= (22/7 × 100) cm 2
Area of circle = 314 cm2
(ii) the shaded triangle
Past cartoon a diagonal for he square, it divides the square into two right triangles.
Area of triangle = one/ii ⋅ b ⋅ h
= 1/2 ⋅ x ⋅ ten
= 50 cm2
(iii) the foursquare
length of diagonal = xx cm
Expanse of square = 1/ii . dtwo
= 1/ii × 20 × xx
Area of foursquare = 200 cm ii
= 200 cmtwo
(b) Percentage = (Expanse of square/Area of circle) × 100
= (200/314) × 100
= 0.6369 × 100
= 63.69%
So, 63.7% of circle is occupied by the foursquare.
Example half dozen :
(a) Find the shaded area.
(b) Find the sum of the perimeters of the two circles.
Solution :
(a) Area of shaded region
= Area of big circumvolve – Expanse of small circle
Let "R" and "r" be the radius of large and pocket-sized circles.
R = 10/2 ==> v m
r = 5/2
Surface area of shaded region = Area of large circumvolve - Expanse of small circumvolve
= ∏R two -∏r 2
=∏(25-(5/2)two)
= ∏(25-(25/four) )
= (22/7) (75/4)
Area of shaded region = 58.92 m2
b) Sum of the perimeters of the two circles
= Perimeter of large circle + Perimeter of small circle
= two∏R+ 2∏r
= two∏(R+r)
= two(22/7) (5+five/two)
= (22/7)(15)
= 47.fourteen m
The sum of perimeter of circles is 47.14 1000.
Example 7 :
A circular carpet is laid on a tiled flooring. Discover
a) the area of the rug
b) the visible expanse of the tiled floor.
Solution :
Radius r = 2.4/2 ==> one.2 m
Area of carpeting = ∏r2
= ∏ × 1.2 × one.ii
= (22/7 × 1.44) m2
= 4.525 m2
Area of carpeting = 4.52 mtwo
b) the visible surface area of the tiled flooring
= Expanse of tiled flooring – Area of rug
Length = 5.2 m, Width = 3.v m
= (5.2 × iii.5) - 4.52
= 18.2-4.52
= 13.68 mtwo
And then, the visible area of the tiled floor is 13.68 m2
Instance 8 :
A gardener is making a path using viii cylindrical concrete pavers. Each paver has a radius of 20 cm and is 5 cm thick.
(a) Detect the total surface area of the tops of the pavers.
(b) Detect the total volume of the pavers
Solution :
Radius of pavers (r) = 20 cm
r = (20/100) k
r = 0.2 m
Expanse of top viii pavers = 8(∏rtwo)
= 8(22/7 × 0.2 × 0.two) mtwo
= 8(0.125) yardtwo
= 1.005 m2
= 1.01 mtwo
Total area of the tops of the pavers is i.01 grand2.
b) r = 0.2 1000
pinnacle (h) = 5 cm ==> 0.05 yard
Volume of the pavers = eight(∏riih)
= 8(22/7 × 0.ii × 0.2 × 0.05)
= 8(0.044/7)
= 8(0.006285) m3
= 0.0503 m3
Book of the pavers = 0.0503 thousand3
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