问题1.如图所示,地块为在BC上具有半圆的矩形ABCD的形式。如果AB = 60 m且BC = 28 m,请找到绘图区域。
解决方案:
Given that, ABCD is a rectangle
So, AB = CD = 60 m
And, BC = AD = 28 m
So, the radius of the semi-circle = BC/2 = 28/2 = 14 m
Now we find the area of the plot = Area of rectangle ABCD + Area of semi – circle
= (l x b) + 1/2 πr2
= (60 x 28) + 1/2 (22/7)(14)2
= 1680 + 308
= 1988 cm2
Hence, the area of the plot is 1988 cm2
问题2。运动场的形状为矩形,在其外侧增加了两个较小直径的半圆。如果矩形的边分别为36 m和24.5 m,请找到操场的区域。 (取π= 22/7)。
解决方案:
Given that,
Length of rectangle = 36 m
Breadth of rectangle = 24.5 m
So, the radius of the semi-circle = breadth/2 = 24.5/2 = 12.25 m
Now we find the area of the playground = Area of the rectangle + 2 x area of semi-circles
= l x b + 2 x 1/2 (πr2)
= (36 x 24.5) + (22/7) x (12.25)2
= 882 + 471.625 = 1353.625
Hence, the area of the playground is 1353.625 m2
问题3。找到圆圈上刻有面积为64 cm 2的正方形的面积。 (使用π= 3.14)
解决方案:
Given that,
Area of square inscribed the circle = 64 cm2
Side2 = 64
So, Side = 8 cm
So, AB = BC = CD = DA = 8 cm
Now In triangle ABC,
By using Pythagoras theorem, we get
AC2 = AB2 + BC2
AC2 = 82 + 82
AC2 = 64 + 64 = 128
AC = √128 = 8√2 cm
So, the angle B = 90o and AC being the diameter of the circle
and the radius is AC/2 = 8√2/2 = 4√2 cm
Now we find the area of the circle = πr2 = 3.14(4√2)2
= 100.48 cm2
Hence, the area of the circle is 100.48 cm2
问题4.一块矩形长20 m,宽15 m。从它的四个角开始,切掉了半径为3.5 m的象限。找到剩余部分的面积。
解决方案:
Given,
Length of the rectangle = 20 m
Breadth of the rectangle = 15 m
Radius of the quadrant = 3.5 m
Now we find the area of the remaining part = (Area of the rectangle) – 4 x( Area of one quadrant)
= (l x b) – 4 x (1/4 x πr2)
= (l x b) – πr2
= (20 x 15) – (22/7)(3.5)2
= 300 – 38.5
= 261.5 m2
Hence, the area of the remaining part is 261.5 m2
问题5.在图中,PQRS是边长4厘米的正方形。找到阴影正方形的区域。
解决方案:
From the above figure we know that, each quadrant is a sector of 90o in
a circle of 1 cm radius. In other words its 1/4th of a circle.
So, its area = 1/4 x πr2
= 1/4 x (22/7)(1)2 = 22/28 cm2
And, the area of the square = side2
It is given that the side = 4 cm
So, 42 = 16 cm2
The area of the circle = πr2 = π(1)2 = 22/7 cm2
Now we find the area of the shaded region = area of the square – area of the circle – 4 x area of a quadrant
= 16 – 22/7 – (4 x 22/28)
= 16 – 22/7 – 22/7 = 16 – 44/7
= 68/7 cm2
Hence, the area of the shaded region is 68/7 cm2
问题6:四头母牛系在50 m边的方形图的四个角上,因此它们无法相互伸手。哪个区域将保持清洁状态? (见图)
解决方案:
Given,
Side of square plot = 50 m
Radius of a quadrant = 25 m
Now we find the area of plot left un-grazed = Area of the plot – 4 x (area of a quadrant)
= Side2 – 4 x (1/4 x πr2)
= (50)2 – 22/7 x (25)2
= 2500 – 1964.28
= 535.72 m2
Hence, the area of plot left un-grazed is 535.72 m2
问题7.在尺寸为20 mx 16 m的矩形区域的拐角处,用长14 m的绳子绑住一头母牛,找到母牛可以放牧的区域。
解决方案:
The dotted portion is the area over which the cow can graze.
So, from the above figure, the shaded area is the area of a quadrant
of a circle of radius equal to the length of the rope.
According to the question it is given that the length of the rope is 14m
So, the radius of the circle is 14 cm
Now we find the required area = 1/4 x πr2
= 1/4 x 22/7 x (14)2
= 154 cm2
Hence, the area of the field in which the cow can graze is 154 cm2
问题8.小牛用长6 m的绳子绑在侧面20 m的方形草皮草坪的拐角处。如果绳索的长度增加了5.5 m,则发现小牛可以放牧的草地的面积增加。
解决方案:
Given,
The initial length of the rope = 6 m
Then the rope is said to be increased by 5.5m
So, the increased length of the rope = (6 + 5.5) = 11.5 m
We know that, the corner of the lawn is a quadrant of a circle.
Now we find the required area = 1/4 x π(11.5)2 – 1/4 x π(5.5)2
= 1/4 x 22/7 (11.52 – 62)
= 1/4 x 22/7 (132.25 – 36)
= 1/4 x 22/7 x 96.25
= 75.625 cm2
Hence, the increase in area of the grassy lawn in which the calf can graze is 75.625 cm2
问题9.方形水箱的边长等于40 m。周围有四个半圆形的草地。以每平方米1.25卢比的价格查找地块的草皮成本(取π= 3.14)。
解决方案:
Given,
Side of the square tank = 40 m
And, the diameter of the semi-circular grassy plot = side of the square tank = 40 m
So, the radius of the grassy plot = 40/2 = 20 m
Now we find the area of the four semi-circular grassy plots = 4 x 1/2 πr2
= 4 x 1/2 (3.14)(20)2
= 2512 m2
Rate of turfing the plot = Rs 1.25 per m2
So, the cost for 2512 m2 = (1.25 x 2512) = Rs 3140
Hence, the area of the four semi-circular grassy plots is 2512 m2
and the cost of turfing the plot at 1.25 per square metre is Rs 3140.
问题10.一个矩形公园是100 m x 50 m。它四周环绕着半圆形的花坛。查找将半圆形花床平整为每平方米60派萨的成本。 (使用π= 3.14)。
解决方案:
Given,
Length of the park = 100 m and the breadth of the park = 50 m
The radius of the semi-circular flower beds = half of the corresponding side of the rectangular park
So, the radius of the bigger flower bed = 100/2 = 50 m
and radius of the smaller flower bed = 50/2 = 25 m
Now we find the total area of the flower beds = 2[Area of bigger flower bed + Area of smaller flower bed]
= 2[1/2 π(50)2 + 1/2 π(25)2]
= π[(50)2 + (25)2]
= 3.14 x [2500 + 625]
= 9812.5 m2
So, rate of levelling flower bed = 60 paise per m2
The total cost of levelling = 9812.5 x 60 = 588750 paise
= Rs 5887.50
Hence, the cost of levelling the semi-circular flower beds at 60 paise per square metre is Rs 5887.50
问题11.一条运行轨道的内周长(如图所示)为400 m。每个笔直部分的长度为90 m,两端为半圆形。如果轨道到处都是14 m宽,请找到轨道的区域。另外,找到外部运行轨道的长度。
解决方案:
Let us considered the radius of the inner semi-circle = r
And that of the outer semi-circle = R
According to the question
The length of the straight portion = 90 m
Width of the track = 14 m
The inner perimeter of the track = 400 m
So, the Inner perimeter of the track = BF + FRG + GC + CQB = 400
90 + πr + 90 + πr = 400
2 πr + 180 = 400
2 x 22/7 x r = 220
r = 35 m
So, the radius of the outer semi-circle = 35 + 14 = 49 m
Now we find the area of the track = 2[Area of the rectangle AEFB + Area of semi-circle APD – Area of semi-circle BQC]
= 2[90 x 14 + 1/2 x 22/7 x 492 – 1/2 x 22/7 x 352]
= 2[1260 + 11 x 7 x 49 – 11 x 5 x 35]
= 2 [1260 + 3773 – 1925]
= 3 x 3108
= 6216 m2
So, the length of outer running track = AE + APD + DH + HSE
= 90 + πR + 90 + πR
= 180 + 2 πR
= 180 + 2 x 22/7 x 49
= 180 + 308
= 488 m
Hence, the area of the track is 6216 m2 and the length of the outer running track is 488 m.
问题12.找到以平方厘米为单位的图形区域,校正到小数点后一位。 (取π= 22/7)。
解决方案:
The radius of the semi-circle = 10/2 = 5 cm
From the above figure,
The area of figure = Area of square + Area of semi-circle – Area of triangle AED
= 10 x 10 + 1/2 πr2 – 1/2 x 6 x 8
= 100 + 1/2 (22/7)(5)2 – 24
= (700 + 275 – 168)/7
= (807)/7
= 115.3 cm2
Hence, the area of the figure, in square cm is 115.3 cm2
问题13。在图中,从AB = 20 cm的矩形区域ABCD中切出AE = 9 cm和DE = 12 cm的直角三角形AED。另一方面,以BC为直径,在该区域的外部添加一个半圆。找到阴影区域的面积。 (使用π= 22/7)。
解决方案:
Given,
Length of the rectangle ABCD = 20 cm
AE = 9 cm and DE = 12 cm
So, the radius of the semi-circle = BC/ 2 or AD/2
Now In triangle AED,
By using Pythagoras theorem, we get
AD = √(AE2 + ED2) = √(92 + 122)
= √(81 + 144)
= √(225) = 15 cm
So, we know that
The area of the rectangle = 20 x 15 = 300 cm2
And, the area of the triangle AED = 1/2 x 12 x 9 = 54 cm2
So, the radius of the semi-circle = 15/2 = 7.5 cm
Area of semi-circle = 1/2 π(15/2)2 = 1/2 x 3.14 x 7.52 = 88.31 cm2
Now we find the area of the shaded region = Area of the rectangle ABCD + Area of semi-circle – Area of triangle AED
= 300 + 88.31 – 54
= 334.31 cm2
Hence, the area of the shaded region is 334.31 cm2
问题14:从一个边长为8 cm的正方形的两个相对角中的每一个切出一个半径为1.4 cm的圆的象限。如图所示,还从中心切出了另一个直径为4.2厘米的圆。找到正方形剩余(阴影)部分的面积。 (使用π= 22/7)。
解决方案:
Given,
Side of the square = 8 cm
Radius of circle = 4.2 cm
Radius of the quadrant = 1.4 cm
Now we find the area of the shaded potion = Area of square – Area of circle – 2 x Area of the quadrant
= side2 – πr2 – 2 x 1/2 πr2
= 82 – π(4.2)2 – 2 x 1/2 π(1.4)2
= 64 – 22/7(4.2 x 4.2) – 22/7(1.4 x 1.4)
= 64 – 388.08/7 – 21.56/7
= 5.48 cm2
Hence, the area of the shaded potion is 5.48 cm2
问题15。在图中,ABCD是一个长方形,AB = 14厘米,BC = 7厘米。以DC,BC和AD为直径,绘制三个半圆,如图所示。找到阴影区域的面积。
解决方案:
Given that,
ABCD is a rectangle with AB = 14 cm and BC = 7 cm
From the above figure, we find
The area of shaded region = Area of rectangle ABCD + 2 x area of semi-circle with AD and BC as diameters – area of the semi-circle with DC as diameter
= 14 x 7 + 2 x 1/2 π(7/2)2 – 1/2 π(7)2
= 98 + 22/7 x (7/2)2 – 1/2 (22/7)(7)2
= 98 + 77/2 – 77
= 59.5 cm2
Hence, the area of the shaded region is 59.5 cm2
问题16。在图中,ABCD是一个矩形,AB = 20 cm,BC = 14 cm。两个180°的扇区已被切除。计算 :
(i)阴影区域的面积。
(ii)阴影区域边界的长度。
解决方案:
Given that,
Length of the rectangle = AB = 20 cm
Breadth of the rectangle = BC = 14 cm
(i) Now we find the area of the shaded region = Area of rectangle – 2 x Area of the semi-circle
= l x b – 2 x 1/2 πr2
= 20 x 14 – (22/7) x 72
= 280 – 154
= 126 cm2
(ii) Now we find the length of the boundary of the shaded region = 2 x AB + 2 x circumference of a semi-circle
= 2 x 20 + 2 x πr
= 40 + 2 x (22/7) x 7
= 40 + 44
= 84 cm
Hence, the area of the shaded region is 126 cm2 and the length of the boundary of the shaded region is 84 cm
问题17。在图中,正方形ABCD分为五个相等的部分,所有部分均具有相同的面积。中心部分是圆形的,线AE,GC,BF和HD沿着正方形的对角线AC和BD放置。如果AB = 22厘米,请找到:
(i)中央部分的周长。
(ii)ABEF部分的周长。
解决方案:
Given that,
Side of the square = 22 cm = AB
Let us considered the radius of the centre part be r cm.
So, the area of the circle = 1/5 x area of the square
πr2 = 1/5 x 222
22/7 x r2 = (22 x 22)/ 5
So, the radius(r) of the circle is = 154/5 = 5.55 cm
(i) Circumference of central part = 2πr = 2(22/7)(5.55) = 34.88 cm
(ii) Let O be the centre of the central part.
Then, its clear that O is also the centre of the square as well.
Now In triangle ABC,
By using Pythagoras theorem
AC2 = AB2 + BC2 = 222 + 222 = 2 x 222
AC = 22√2
AO = 1/2 AC = 1/2 (22√2) = 11√2 cm [Since diagonals of a square bisect each other]
And,
AE = BF = OA – OE = 11√2 – 5.55 = 15.51 – 5.55 = 9.96 cm
EF = 1/4 (Circumference of the circle) = 2πr/4
= 1/2 x 22/7 x 5.55 = 8.72 cm
Now we find the perimeter of the part ABEF = AB + AE + EF + BF
= 22 + 2 x 9.96 + 8.72
= 50.64 cm
Hence, the Circumference of central part is 34.88 cm and
the perimeter of the part ABEF is 50.64 cm