第12章与圆相关的区域–练习12.3 |套装1
问题11.在方形手帕上，制作了九个圆形设计，每个圆形设计的半径为7厘米(见图)。找到手帕剩余部分的面积。
解决方案：
Side of square=6*Radius
=6*7
=42 cm
Area of shaded region=Area of square – Area of 9 circles
=side*side – 9πr2
=42*42 – 9*22/7*7*7
=1764-1386
=378 cm2
The area of the remaining portion of the handkerchief =378 cm2
问题12。在图中，OACB是中心为O，半径为3.5 cm的圆的象限。如果OD = 2厘米，请找到
(i)象限OACB，(ii)阴影区域。
解决方案：
(i) Area of shaded region=Area of quadrant
=1/4πr2
=1/4*22/7*3.5*3.5
=38.5/4
=9.625 cm2
(ii) Area of shaded region=Area of quadrant -Area of ∆BOD
=9.625-1/2*BO*OD
=9.625-1/2*3.5*2
=9.625-3.5
=6.125 cm2
问题13。在图中，正方形OABC刻在象限OPBQ中。如果OA = 20厘米，请找到阴影区域。 (使用π = 3.14)
解决方案：
By Pythagoras theorem,
OB2=DA2+AB2
BO2=(20)2+(20)2
OB2=400+400
OB2=800
OB=√800
OB=√(2*2*2*2*5*5)
OB=2*2*5√2
OB=20√2
Area of shaded region=Area of quadrant -Area of square
=1/4πr2 – side*side
=1/4*3.14*20√2*20√2-20*20
=1/4*3.14*800-400
=1*3.14
=22cm2
=1/4*3.14
The area of shaded region is =1/4*3.14
问题14. AB和CD分别是半径为21 cm和7 cm的两个同心圆和中心为O的圆弧(见图)。如果∠AOB= 30°，请找到阴影区域的面积。
解决方案：
Area of shaded region=Area of sector AOB-Area of sector COD
=θ/(360°) πR2-θ/(360°) πr2
=θ/(360°) π[R2-r2]
=30°/360*22/7[(21)2-(7)2]
=1/12*22/7*28*14
=308/3cm2
=102.66cm2
The area of shaded region 102.66cm2
问题15。在图中，ABC是半径为14 cm的圆的象限，并以BC为直径绘制了一个半圆。找到阴影区域的面积。
解决方案：
Area of segment=Area of quadrant -Area of ∆BAC
=1/4πr2-1/2*AC*AB
=1/4*22/7*14*14-1/2*14*14
=11*14-98
=154-98
=56 cm2
Semicircle R=?
In rt. ∆BAC, By Pythagoras theorem,
BC2=AB2+BC2
BC2= (14)2+(14)2
BC=√((14)2+(14)2)
BC=√((14)2[1+1] )
BC=14√2
∴Diameter of semicircle=14√2cm
then radius R of semicircle=14√2/2=7√2cm
Area of semicircle =1/2πR2
=1/2*22/7*7√2*7√2
=22*7
=154 cm2
Area of shaded region=Area of semicircle-Area of segment
=154-56 cm2
=98 cm2
The area of shaded region is =98cm2
问题16.计算图中设计区域的面积，该区域在两个半径均为8 cm的圆的两个象限之间共有。
解决方案：
Area of design=Area of 2 quadrant -Area of square
=2*1/4πr2-side*side
=1/2*22/7*8*8-8*8
=704/7-64
=100.57-64
=36.57cm2
Area designed region in figure is 36.57cm2
