问题18.在给定的图中,找到阴影区域的面积。 (使用π= 3.14)。
解决方案:
The side of the square = 14 cm
So, area = side2
142 = 196 cm2
Let’s assume the radius of each semi-circle be r cm.
Then,
r + 2r + r = 14 – 3 – 3
4r = 8
r = 2
So, the radius of each semi-circle is 2 cm.
Area of 4 semi-circles = (4 x 1/2 x 3.14 x 2 x 2) = 25.12 cm2
Now,
the length of side of the smaller square = 2r = 2 x 2 = 4 cm
So, the area of smaller square = 4×4 = 16 cm2
Now we find the area of unshaded region = Area of 4 semi-circles + Area of smaller square
= (25.12 + 16) = 41.12 cm2
Now we find the area of shaded region = Area of square ABCD – Area of unshaded region
= (196 – 41.12) = 154.88 cm2
Hence, the area of the shaded region 154.88 cm2
问题19.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:
Given that,
Radius of small quadrant, r = 2 cm
Radius of big quadrant, R = 3.5 cm
(i) Area of quadrant OACB = 1/4 πR2
= 1/4 (22/7)(3.5)2
= 269.5/28 = 9.625 cm2
(ii) Area of shaded region = Area of big quadrant – Area of small quadrant
= 1/4 π(R2 – r2)
= 1/4 (22/7)(3.52 – 22)
= 1/4 (22/7)(12.25 – 4)
= 1/4 (22/7)(8.25)
= 6.482 cm2
Hence, the area of the quadrant OACB is 9.625 cm2 and shaded region is 6.482 cm2
问题20.在图中,正方形OABC刻在圆的象限OPBQ中。如果OA = 21厘米,请找到阴影区域的面积。
解决方案:
Given,
Side of the square = 21 cm = OA
Area of the square = OA2 = 212 = 441 cm2
Diagonal of the square OB = √2 OA = 21√2 cm
From the figure its seen that,
The diagonal of the square is equal to the radius of the circle, r = 21√2 cm
So,
The area of the quadrant = 1/4 πr2 = 1/4 (22/7)(21√2)2 = 693 cm2
Now we find the area of the shaded region = Area of the quadrant – Area of the square
= 693 – 441
= 252 cm2
Hence, the area of the shaded region is 252 cm2
问题21.在图中,OABC是边长7厘米的正方形。如果OAPC是中心为O的圆弧象限,则找到阴影区域的面积。 (使用π= 22/7)
解决方案:
Given that,
Side of OABC square = 7 cm
So, OA = AB = BC = OC = 7 cm
Now, the area of square OABC = side2 = 72 = 49 cm2
It is given that OAPC is a quadrant of a circle with centre O.
So, the radius of the quadrant = OA = OC = 7 cm
Area of the OAPC quadrant = 90/360 x πr2
= 1/4 x (22/7) x 72
= 77/2 = 38.5 cm2
Now we find the area of shaded portion = Area of square OABC – Area of quadrant OAPC
= (49 – 38.5) = 10.5 cm2
Hence, the area of shaded portion is 10.5 cm2
问题22。在图中,OE = 20厘米。在OSFT扇区中,刻有方形OEFG。找到阴影区域的面积。
解决方案:
Given that,
Side of OEFG square = 20 cm.
So its diagonal = √2 side = 20√2 cm
And, the radius of the quadrant(r) = diagonal of the square
r = 20√2 cm
Now we find the area of the shaded portion = Area of quadrant – Area of square
= 1/4 πr2 – side2
= 1/4 (22/7)(20√2)2 – (20)2
= 1/4 (22/7)(800) – 400
= 400 x 4/7 = 1600/7 = 228.5 cm2
Hence, the area of the shaded region 228.5 cm2
问题23.如果AC = 24 cm,BC = 10 cm,O是圆心,则找到图中阴影区域的面积。 (使用π= 3.14)
解决方案:
Given,
AC = 24 cm and BC = 10 cm
From the figure AB is the diameter of the circle
So, ∠ACB = 90o
By using Pythagoras theorem
AB2 = AC2 + BC2 = 242 + 102 = 576 + 100 = 676
So, AB = √676 = 26 cm
So, the radius of the circle(r) = 26/2 = 13 cm
Now we find the area of shaded region = Area of semi-circle – Area of triangle ACB
= 1/2 πr2 – 1/2 x b x h
= 1/2 (22/7)132 – 1/2 x 10 x 24
= 265.33 – 120
= 145.33 cm2
Hence, the area of shaded region is 145.33 cm2
问题24.在等边三角形ABC的边上刻有一个12厘米的圆形,接触其侧面(见图)。找到内接圆的半径和阴影部分的面积。
解决方案:
Given that,
Side of ABC triangle = 12 cm
So, the area of the equilateral triangle = √3/4(side)2
= √3/4(12)2 = 36√3 cm2
Also the perimeter of triangle ABC = 3 x 12 = 36 cm
Now we find the radius of incircle = Area of triangle/ ½ (perimeter of triangle)
= 36√3/ 1/2 x 36
= 2√3 cm
Also, we find the area of the shaded part = Area of equilateral triangle – Area of circle
= 36√3 – πr2
= 36(1.732) – (3.14)(2√3)2
= 62.352 – 37.68
= 24.672 cm2
Hence, the area of the shaded part is 24.672 cm2
问题25.在图中,边长为6厘米的等边三角形ABC刻在一个圆上。找到阴影区域的面积。 (取π= 3.14)
解决方案:
Given that,
The side of ABC triangle = 6 cm
So, the area of the equilateral triangle = √3/4(side)2
= √3/4(6)2
= √3/4(36)
= 9√3 cm2
Let us mark the centre of the circle as O, OA and OB are the radii of the circle.
In triangle BOD,
sin 60o = BD/ OB
√3/2 = 3/ OB
OB = 2√3 cm = r
Now we find the area of shaded region = Area of the circle – area of the equilateral triangle
= πr2 – 9√3
= 3.14 x (2√3)2 – 9√3
= 3.14 x 12 – 9 x 1.732
= 37.68 – 15.588
= 22.092 cm2
Hence, the area of the shaded region 22.092 cm2
问题26.圆形场的周长为650 m。在该字段中标记一个正方形图,该图的顶点在该字段的圆周上。计算平方图的面积。
解决方案:
Given that, the circular field has a perimeter = 650 m
As we know that the circumference of circle = 2πr
650 = 2×22/7xr
r = (650×7)/44
r = 103.409 m
As the diagonal of the square plot is the diameter of the circle.
Hence, r × 2 = d
So, diameter = 103.409 x 2
As we know that the Diameter of circle = 206.818 m = Diagonal of square plot
Now we find the area of the square plot = 1/2 × d²
= 1/2 × (206.818)²
= 1/2 × 42773.68
Hence, the area of the square plot is 21386.84 m2
问题27.在图中找到阴影区域的区域,其中已绘制了半径7 cm的圆弧,以边长14 cm的等边三角形ABC的顶点A为中心。 (使用π= 22/7和√3= 1.73)
解决方案:
Given that,
The radius of circle (r) = 7 cm.
Side of the triangle = 14 cm
Now we find the area of shaded region = Area of the circle + Area of the triangle – 2 area of the sector EAF
= πr2 + √3a2/4 – 2 × πr2 × 60/360
= 22/7 × (7)2 + (1.73)(14)2/4 – 2 × 22/7 × (7)2 × 60/360
= 283.77 – 51.33
= 187.43cm2
Hence, the area of the shaded region is 187.43cm2
问题28.正六边形被刻成一个圆。如果六边形的面积为24√3cm 2 ,则找到圆的面积。 (使用π= 3.14)
解决方案:
Given that,
A regular hexagon ABCDEF is inscribed in a circle
Area of hexagon = 24 √3 cm2
Let us considered r be the radius of circle
So, the side of regular hexagon = r
Area of equilateral ΔOAB = √3/3 r2 cm2
But the area of triangle OAB = 1/6 x area of hexagon
24√3/6 = 4√3 r2
So, r2 = 4√3 x 4/√3 = 16
r = 4 cm
Now we find the area of circle = πr2 = 3.14 x (4)2 cm2
= 3.14 x 16 cm2
= 50.24 cm2
Hence, the area of circle = 50.24 cm2
问题29. ABCDEF是中心为O的正六边形(见图)。如果三角形OAB的面积为9 cm 2 ,则找到以下面积:
(i)六角形和
(ii)内切六边形的圆。
解决方案:
The area of hexagon = 54 cm2
and area of circle = 65.324 cm2
Given that,
Area of ∆OAB = 9 cm2
Let us considered ‘r’ be the radius of a circle and ‘a’ be the side of a equilateral triangle.
(i) As we know that the area of hexagon = 6 × Area of equilateral triangle
= 6 × 9 = 54 cm²
= 54 cm²
(ii) Area of equilateral ∆OAB = √3/4 × a2
9 = √3/4 × a2
9 × 4 = √3a2
36 = √3a2
a2 = 36/√3
Side2 = 36/√3 cm
As we know that in regular hexagon inscribed in a circle,
its side is equal to the radius of a Circle
So, the radius of circle(r) = Side of a hexagon
r2 = 36/√3 cm
Now we find the area of circle = πr2
= 22/7 × 36/√3
= 22/7 × 36/1.732
= (22 × 36) /(7 ×1.732)
= 792/12.124
= 65.324 cm2
Hence, the area of hexagon is 54 cm2 and area of circle is 65.324 cm2
问题30.四个相等的圆,每个圆的半径均为5 cm,如图所示。找到它们之间包括的区域。 (取π= 3.14)。
解决方案:
Given that,
Radius of a circle = 5 cm
Side of a square = 2 × Radius of a circle
= 2 × 5
Side of a square = 10 cm
Area of a square = Side2
Area of a square = 102 = 100 cm2
Area of a square = 100 cm2
Area of the quadrant of one circle = 1/4πr2
Area of the quadrant of four circles = 4 × 1/4πr2 = πr2
= 3.14 × 5 × 5
= 3.14 × 25
= 78.5cm²
Area of the quadrant of four circles = 78.5cm2
Now we find the area of the shaded portion = Area of the square – Area of the quadrant of four circles
= 100 – 78.5
= 21.5 cm2
Hence, the area of the shaded portion is 21.5 cm2.
问题31.四个相等的圆,每个圆的半径为’a’彼此接触。表明,它们之间的区域是图6a7分之2。 (取π= 22/7)
解决方案:
Given that,
The radius of a circle = a
Side of a square = 2 × Radius of a circle
= 2 × a
Side of a square = 2a cm
Area of a square = Side2
= (2a)2
= 4a2
Area of the quadrant of one circle = 1/4πr2
Area of the quadrant of four circles = 4 × 1/4πr2 = πr2
Now we find the area of the shaded portion = Area of the square – Area of the quadrant of four circles
= 4a2 – 22/7 × a2
= 4a2 – 22a2/7
= (28a2 -22a2)/7
= 6a2/7
Hence, proved the area between them is 6a2/7
问题32.一个孩子在一张纸上画一张海报,画出一个边长14厘米的正方形ABCD。她用中心A,B,C和D绘制了四个圆圈,在其中提出了不同的节能方式。绘制圆圈的方式应使每个圆圈从外部接触给定图中三个剩余的圆圈中的两个。她在阴影区域写了一条消息“节省能源”。找到阴影区域的周长和面积。 (使用π= 22/7)
解决方案:
Given that,
Side of a square = 14 cm
Radius of a Circle(r) = Side of a square/2 = 14/2 = 7 cm
Central angle, θ = 90°
So, the perimeter of the shaded portion = 4 × length of the arc having Central angle 90°
= 4 × θ/360° × 2πr
= 4 × 90°/360° × 2 × 22/7 × 7
= 4 × 1/4 × 44
= 44 cm
As we know that area of a square = Side2
= (14)2 = 196 cm2
Also, the area of a square = 196 cm2
Area of the quadrant of one circle = 1/4πr2
Area of the quadrant of four circles = 4 × 1/4πr2 = πr2
= 22/7 × 72
= 22 × 7
= 154 cm2
Now we find the area of the shaded portion = Area of the square ABCD – Area of the quadrant of four circles
= 196 – 154
= 42 cm2
Hence, the Perimeter of the shaded portion is 44 cm and area of the shaded portion is 42 cm2
问题33.硬币的直径为1厘米(见图)。如果将四个这样的硬币放在桌子上,使每个硬币的边缘与其他两个硬币的边缘接触,则找到阴影区域的面积(取π= 3.1416)。
解决方案:
Given that,
The diameter of a coin (circle) = 1 cm
Find: the area of the shaded region
As we know that the radius(r) of a coin (circle) = Diameter of a coin (circle)/2
r = 1/2 = 0.5 cm
Side of a square = 2 × Radius of a coin (circle)
= 2 × 0.5
= 1 cm
Area of the quadrant of one circle = 1/4πr2
Area of the quadrant of four circles = 4 × 1/4πr2 = πr2
Now we find the area of the shaded portion = Area of the square – Area of the quadrant of four circles
A = Side2 – πr2
= 12 – 3.1416 × (0.5)2
= 1 – 3.1416 × 0.25
= 1 – 0.7854
= 0.2146 cm2
Hence, the area of the shaded portion is 0.2146 cm2
问题34.从尺寸为14厘米x 7厘米的矩形纸板中切出彼此接触的两个相等半径和最大面积的圆形片。找到剩余的卡板区域。 (使用π= 22/7)
解决方案:
Given that,
Length of a rectangle(l) = 14 cm
Breadth of a rectangle(b) = 7 cm
So, the diameter of 1 circle = 7 cm
Radius of 1 circle(r) = 7/2 cm
Now we find the area of the remaining cardboard = Area of rectangular cardboard – 2 × Area of circle
= l × b – 2 (πr2)
= (14× 7) – 2 × (22/7) × (7/2)2
= 98 – (44/7)(49/4)
= 98 – 77
= 21 cm2
Hence, the area of the remaining cardboard is 21 cm2.