10类NCERT解决方案-第12章与圆相关的区域–练习12.2 |套装1

问题1.如果扇形的角度为60°，则找到半径为6 cm的圆形扇形的面积。

解决方案：

Given: r=66cm and ∅=60°

_{Area of sector=∅/(360°) *πr}2

=60/360*22/7*6*6

=132/7

问题2。找到圆周为22 cm的圆的象限的面积。圆的象限是扇形成90°。

解决方案：

Given: circumference of circle=22cm and ∅=90°

To find r=?

2πr=22

Radius =r=22/2πr cm=7/2cm

_{Area of quadrant =∅/(360°) *πr}2

=90/360*22/7*7/2*7/2

=77/8cm2

问题3.时钟分针的长度为14厘米。在5分钟内找到分针扫过的区域。

解决方案：

Let assume, Minute hand of clock acts as radius of the circle.

Angle rotated by min⁡ (hand in 5minutes)=∅=360/60*5=30°

Radius=r=14cm

Area of swept middle hand=

_{=∅/(360°) *πr}2

=30/360*22/7*14*14

_=154/3cm2

_{Area swept by the minute hand in 5min=154/3cm}2

问题4：一个半径为10 cm的圆的弦在中心直角对齐。找到相应的区域：(i)次要部门(ii)主要部门。 (使用π = 3.14)

解决方案：

Radius =r=10cm

Major segment is =360°-90°=270°

(i) Area of minor segment =Area of sector-Area of triangle

=∅/(360°) *πr2-1/2*h*b

=90/306*3.14*10*10-1/2*10*10

=314/4-50

=78.5-50

_{Area of minor segment =28.5cm}2

_{(ii) Area of major sector=∅/(360°) *πr}2

=270/360*3.14*10*10

_{=3*314/4=235.5cm}2

_{Area of major segment=235.5cm}2

问题5：在半径为21 cm的圆中，圆弧在中心处对角为60°。找：

(i)弧长

(ii)弧线所形成的扇形区域

(iii)由相应的和弦形成的线段区域

解决方案：

(i) Radius=r=21cm

Length of arc=∅/(360°) *2πr

60/360*2*22/7*21

=22cm

_{(ii) Area of sector=∅/(360°) *πr}2

60/360*22/7*21*21

_11*21=231cm2

(iii) Area of segment =Area of sector -Area of triangle

_{=231-√3/4(side)}2

=231-1.73/4*21*21

=231-762.93/4

=231-190.73

_=40.27cm2

问题6.半径为15 cm的圆的弦在中心对角为60°。找到圆的相应次要和主要部分的区域。 (使用π= 3.14和√3= 1.73)

解决方案：

Radius of circle=15cm

∆AOB is isosceles

∴∠A = ∠B

=∠A+∠B+C=180°

=2∠A=180°-60°

=∠A=120°/2

=∠A=60°

Area of minor segment =Area of sector-Area of triangle

_{=∅/(360°) *πr}2-_√3/4(side)2

=(60°)/360*3.14*15*15-1.73/4

=706.5/6-389.25/4

=117.75-97.31

=_20.44cm2

Area of major segment-Area of circle-Area of minor segment

_=πr2-20.44

=3.14*15*15-20.44

_=686.06cm2

问题7.半径为12 cm的圆的弦在中心对角120°。找到圆的相应线段的面积。 (使用π = 3.14和√3= 1.73)

解决方案：

Radius=r=12cm

Area of triangle=1/2*base*height

Area of segment=Area of sector -area of triangle

=∅/(360°)*π*r²-1/2(side)²*sin∅

=120°/360°*3.14-1/2(side)²*sin120°∅

=150.72-6*12*sin60°

=150.72-6*12*√3/2

=150.72-36*1.73

=150.72-62.28

_=88.44cm2

第12章与圆相关的区域–练习12.2 |套装2