Rd Sharma Solutions Class 8 Chapter 20 Mensuration – I Area of a trapezium and a polygon
Welcome to NCTB Solution. Here with this post we are going to help 8th Class students for the Solutions of RD Sharma Class 8 Mathematics, Chapter 20, Mensuration – I (Area of a trapezium and a polygon). Here students can easily find Exercise wise solution for chapter 20, Mensuration – I (Area of a trapezium and a polygon). Students will find proper solutions for Exercise 20.1, 20.2 and 20.3 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily.
Mensuration – I Area of a trapezium and a polygon Exercise 20.1 Solution :
Question no – (1)
Solution :
Area of parallelogram,
= 24 × 10
= 240 cm2
∴ 1080 m2
= 1080 × 100 × 100
= 10800000 cm2
∴ 1080 cm2 area need tiles
= 10800000/240
= 45000
Therefore, 45000 tiles are required to cover a floor of area 1080 m2
Question no – (2)
Solution :
∴ Area of ABCD
= 28 × 60
= 1680 m2
∴ Area of semi-circle,
= πr2/2
= 22 × 28 × 28/7 × 2 × 2 × 2
= 308 m2
∴ Area of ABCD + area of semicircle,
= 1680 + 308
= 1988 m2
Therefore, the area of the plot will be 1988 m2
Question no – (3)
Solution :
Two semicircle are denoted by = C, and C2
∴ Area of ABCD = 36 × 24.5
= 882 m2
Area of C, = πr2/2
∴ Area of 17.5 C2
= 22 × 24.5 × 24.5/7 × 2 × 2
= 235.812
∴ Total playground Area of C + Area of ABCD + Area of c2
= 235.812 + 882 + 235.812
= 1353.624 m2
Therefore, the area of the playground is 1353.624 m2
Question no – (4)
Solution :
First, Area of quadrant,
= 0/360 × πr2
= 90/360 × 22/7 × (3.5)2
= 9.625m2
∴ Area of 4 quadrant,
= 4 × 9.625
= 38.5 00 m2
∴ Area of rectangular piece,
= 20 × 15
= 300 m2
∴ Area of remaining part of rectangular,
= (300 – 38.5)
= 261.5 m2
Question no – (5)
Solution :
As per the question,
The inside perimeter of track = 400 m
Total length two straight portion,
= 90 + 90
= 180 m
According questions,
= 2πr = 220
or, r = 220/2π = 220 × 7/ 2 × 22 = 35 m
Radius of circular portion of outer running track,
= 35 + 14 m
= 49
Area of track (2 × 90 × 14) + 2} 22/7 × 1/2 [(49)2 – (35)2]
= 2520 + 22 ×2/7 × 2 (2401) – 1225)
= 2520 + 22/7 × 1176 = 2520 + 3696
= 6218 m2
∴ Length of the outer running track,
= 180 +2 × 22/7 × 49
= 180 + 380
= 488 m.
Therefore, the length of the outer running track will be 488 m.
Question no – (6)
Solution :
We, connect A and D
∴ AD = 10 cm = CD
∴ Area of △ADE
= 1/2 × 6 × 8
= 24 cm2
∴ Area of ABCD
= 10 × 10
= 100 cm2
∴ Area of semicircle
= πr22 = 1/2 × 22/7 × (10/2)2
= 11/7 × 52 = 11 × 25/7
= 39.28 cm2
∴ Area of ABCDE = area of ABCD – area of △ADE
= 100 – 24
= 76 cm2
∴ Area of given fig = area of (△ ABCDE + semicircle) 39.28
= 76 + 39.28
= 115.28 cm2
Question no – (7)
Solution :
∴ Diameter of wheel = 90 cm
∴ r = 90/2
= 45 cm
∴ Perimeter of wheel,
= 2πr = 2 × 22/7 45
= 1980/7 cm
∴ Distance covered in 315 revolution,
= 1980/7 × 315
= 89100 cm
= 891 m
= 891 m
= 891/1000
= 0.891 km
∴ In 1 min speed will be
= 0.98 km
∴ In 1hr = 60 min speed will be
= 0.891×60
= 53.46 km/hr
So, its speed in kilometers per hour is 53.46 km/hr
Question no – (8)
Solution :
Area of rhombus,
= 240 cm2
= d1 = 16 cm
∴ Let, another diagonal = d2
According to questions,
= 1/2 (d1 × d2) = 240
or, d2 = 240 × 2/d1
= 240 × 2/16
= 30 cm
Therefore, another diagonal will be 30 cm.
Question no – (9)
Solution :
Let, = d1 = 7.5 cm
And, = d2 = 12 cm
∴ Area = 1/2 (d1× d2)
= 1/2 × 7.5 × 12
= 450 cm
Therefore, the area of rhombus will be 450 cm.
Question no – (10)
Solution :
As per the given question,
The diagonal of a quadrilateral shaped field is = 24 m
Remaining opposite vertices are = 8 m and 13 m
∴ Area of field = area of (△ABD + △BCD)
= (1/2 × 24 × 8) + (1/2 × 24 × 13)
= 96 + 156
= 252 cm2
Therefore, the area of the field will be 252 cm2
Question no – (11)
Solution :
d1 = 8 cm
Side of rhombus = 6 cm
Height = 4 cm
∴ Area of 1/2 part of rhombus,
= 1/2 × 6 × 4
= 12 cm2
∴ Total area of rhombus,
= 12 × 2
= 24 cm2
Now, according to questions,
Area of rhombus,
= 1/2 × d1 × d2
or, 24 = 1/2 × 8 × d2
or, d2 = 24 × 2/8
= 6 cm
Therefore, the length of the other diagonal is 6 cm.
Question no – (12)
Solution :
Let, d1 = 45 cm
And, d2 = 30 cm
∴ Area of rhombus, = A = 1/2 × d1 × d2
or, A = 1/2 × 45 × 30
= 675 cm2
= 0.0675 cm2
= 0.0675 floor consists 3000 tiles.
∴ Polish area,
= 0.0675 × 3000
= 202.5 m2
∴ Total cost,
= 202.5 × 4
= 810 Rs
Therefore, the total cost of the polishing the floor will be 810 Rs.
Question no – (13)
Solution :
Total grassy plot area,
= 112 × 78
= 8736 m2
Area if inner grassy plot’s area,
= (112 – 2 × 2.5) × (78 – 2 × 2.5)
= (112 – 5) × (78 – 5)
= 107 × 73
= 7811 m2
∴ Area of gravel path,
= 8736 – 7811
= 925 m2
∴ 1 m2 path constructing cost = 4.50
∴ 925 m2 path constructing cost,
= 925 × 4.50
= 4162.50 Rs.
Therefore, the area of the path will be 925 m2 and total cost will be 4162.50 Rs.
Question no – (14)
Solution :
d1 = 24 cm = AC
∴ AO = OC = 24/2 = 12 cm
AB = BC = CD = AD = 20 CM
∴ △AOD,
(AB)2 = (AO)2 + (BO)2
or, (BO)2 = (AB)2 – (A0)2
= (20)2 – (12)2
= 400 – 144
= 256 cm
or, BO = √256
= 16 cm
∴ BD = d2 = 2 × BO
= 2 × 16
= 32 cm
∴ Area of rhombus,
= 1/2 × d1 × d2 = 1/2 × 24 × 32
= 384 cm2
Therefore, the area rhombus will be 384 cm2
Question no – (15)
Solution :
Let, Altitude of Rhombus = 4
Length of Square field Ls = 4 m
∴ Area of square field = As = 4 × 4= 16 m2
∴ Area of square field = As = AR = Area of rhombus = 16m2
d1 of rhombus = 2 m
∴ d2 of rhombus = ?
Now, AS = 1/2 × d1 × d2
or, d2 = As × 2/r1 = 16 × 2/2 = 16 cm
∴ Side of rhombus,
= 1/2 √d12 + d22 = 1/2 √22 + 16
= 1/2 √4 + 256 = 1/2 √260
= H = area of Rhombus /side =16/1/1/2 √260
= 16/1/1/2 × 2√65
= 16/√65 cm
Therefore, the altitude of the rhombus will be 16/√65 cm
Question no – (16)
Solution :
∴ 1/2 Area of rhombus,
= 1/2 × 14 × 16
= 112 m2
∴ Area of rhombus,
= 112 × 2
= 224 m2
Therefore, the area of the field will be 224 m2
Question no – (17)
Solution :
Fencing square field 1200,
= 1200/600/100
= 1200 × 100/60
= 2000 m
∴ Square field’s perimeter = 2000 m
Let, one side of square field = a
According to questions,
= 4a = 2000
or, a = 2000/4 = 500 m
∴ Area = a2 = 500 × 500 = 250000 m2
∴ Now, 100 square metre rate is = 50 pipe
∴ 1 sqm rate = 50/100 = 1/2 paise
= 1 / 2 × 100 = 1/200
∴ 25000 sqm rate is,
= 250000 × 1/2
= 1250 Rs.
Therefore, the cost of the reaping the field will be 1250 Rs.
Question no – (18)
Solution :
Area of square plot = As = 84 × 84 = 7056 m2
Length of rectangle plot = 144 m
Let, Width rectangle plot = W m
Now, according to question,
As = AR
or, As = 144 × W
or, w = 7056/144
= 49 m
Therefore, the width of the rectangular plot will be 49 m.
Question no – (19)
Solution :
Let, altitude or rhombus = H
According to questions,
= 1/2 of Area of Rhombus = 1/2 × 10 × H
or, 1/2 × 84 = 1/2 × 10 H
or, H = 84 × 2/2 × 10
= 8.4 m
Therefore, its altitude will be 8.4 m
Question no – (20)
Solution :
First, 1/2 of are of garden,
= 1/2 × 30 × 16
= 240 m2
∴ Area of garden,
= 240 × 2
= 480 m2
∴ Cost of 1 m2 of levelling = 2
∴ Cost of 480 m2 levelling,
= 2 × 480
= 960 Rs.
Therefore, the cost of leveling the garden will be Rs. 960
Question no – (21)
Solution :
First, 1/2 Area of rhombus,
= 1/2 × 64 × 16
= 512 m2
∴ Area of rhombus,
= Ar = 512 × 2
= 1024 m2
Area of square = As = Ar = 1024 m2
∴ Side of square
= √1024
= 32 m
Therefore, the side of a square field is 32 m.
Question no – (22)
Solution :
First, the area of triangle,
= (1/2 × 24.8 × 16.5) cm²
= 204.6 cm²
Here, Area of rhombus = Area of triangle
Given, length of one diagonal = 22 cm
Let, other diagonal d²
Therefore,
= 1/2 × 22 × d² = 204.6
= d² = (204.6 × 2/22)
= 18.6 cm
Therefore, the length of the other diagonal is 18.6 cm.
Mensuration – I Area of a trapezium and a polygon Exercise 20.2 Solution :
Question no – (1)
Solution :
(i) bases = 12 dm and 20 dm, altitude = 10 dm
Bases = 12 dm 20 dm
Altitude = 10 dm.
∴ Area = 1/2 × (12 + 20) × 10
= 1/2 × 23 × 10 = 160 dm2
= 1.6 m2
(ii) bases = 28 cm and 3 dm, altitude = 25 cm
Bases = 28 cm = 0.28 m
= 3 dm = 30 m
∴ Area = 1/2 × (0.28 + 30) × 0.25
= 1/2 × 0.58 × 0.25
= 0.725 m2
(iii) bases = 8 m and 60 dm, altitude = 40 dm
Bases = 8 m = 60 dm = 600 m
altitude = 409 dm = 400 m
∴ Area = 1/2 × (8 + 600) × 400
= 1/2 × 608 × 4000
= 121,600 m2
(iv) bases = 150 cm and 30 dm, altitude = 9 dm.
Bases = 150 cm = 1.5 m
= 30 dm = 30000
Altitude = 9 dm = 90 m
∴ Area = 1/2 × (1.5 + 300) × 90
= 1/2 × 301.50 × y90
= 13,567.5 m2
Question no – (2)
Solution :
As per the question,
Base 15 cm and Height 8 cm,
Side parallel to the given base is 9 cm long
∴ Area,
= 1/2 × (15 + 9) × 8
= 1/2 × 24 × 8
= 96 cm2
Therefore, the area of trapezium will be 96 cm2
Question no – (3)
Solution :
According the question,
Length = 16 dm and 22 dm
Height is = 12 dm
∴ Area,
= 1/2 × (16 + 22) × 12
= 1/2 × 38 × 12
= 228 dm2
Therefore, the area of a trapezium will be 228 dm2
Question no – (4)
Solution :
As per the given question,
Bases (parallel sides) is = 60 cm
Area is = 600 cm2
Let, height = H
According to questions,
= 1/2 × H × 60 = – 600
or, H = 600 × 2/60
∴ H = 2.0 cm
Therefore, the height of a trapezium is 2.0 cm.
Question no – (5)
Solution :
According to the given question,
Area is = 65 cm2
Bases are = 13 cm and = 26 cm
Let, the altitude = H
According to questions,
= 1/2 × H × (13 + 26) = 65
or, H/2 × (39) = 65
or, H = 65 × 2/39
∴ H = 10/3 cm
Therefore, the altitude of a trapezium will be 10/3 cm.
Question no – (6)
Solution :
Let, sum of length bases = x
According to questions,
= 1/2 × H × x = 4.2 m2
or, x = 4.2 × 2/2.8 = 3 [∵ H = 28 cm= 2.8 cm]
∴ x = 3 m
Therefore, the sum of the lengths will be 3 m.
Question no – (7)
Solution :
∴ Area of ABPQ = 1/2 × (10 + 15) × 6
= 1/2 × 25 × 6
= 75 cm2
(i) Area of ABCD = 10 × 15 = 150 cm2
∴ Area of (△ ADQ + △BCP)
= (150 – 75)
= 75 cm2
∴ Area of (△ABQ △BCP) + △Area of ABCD
= 75 + 150
= 225 cm2
(ii) Area of ·ABCD – Area of (△ADQ + △BCP)
= 150 – 75
= 75 cm2
Question no – (8)
Solution :
Let, the distance = h
Now, as per the question,
= 1/2 × (24 + 46) × h = 960
or, 80 × h/2 = 960
or, h = 960 × 2/80
= 24 cm
Therefore, the distance between them is 24 cm.
Question no – (9)
Solution :
Height of trapezium,
= 70 – 50
= 20 cm
∴ Area of ·ABHG,
= CDFE = 1/2 × (30 + 10) × 20
= 1/2 × 40 × 20 = 400 cm2
∴ Area of (ABHG + ABDC + CDFE),
= 400 + 500 + 400
= 13000 cm2
Therefore, the area of the figure will be 13000 cm2
Question no – (10)
Solution :
As per the given question,
Its parallel sides are = 1 m and 1.2 m
perpendicular distance between them is = 0.8
∴ Area of table,
= 1/2 × (1 + 1.2) × 0.8
= 1/2 × 2.2 × 0.8
= 0.88 m2
Therefore, the are of the table will be 0.88 m2
Question no – (11)
Solution :
According to the question,
Canal is = 10 m wide at the top = 6 m wide
Bottom and the area of cross-section is = 72 m2
Let, the depth = h
∴ According to questions,
= 1/2 × h = (10 + 6) = 72
or, h × 16 = 72 × 2
or, h = 72 × 2 /16
∴ h = 9 m
Therefore, the depth of the canal will be 9 m.
Question no – (12)
Solution :
Let, one side = x cm
∴ Another side = (x + 8) cm
Now, according to questions,
= 1/2 × [x + (x + 8) ] × = 7 = 91
or, x + x + 8 = 91 × 2/7 = 26
or, 2x = 26 – 8 = 18
or, x = 18/2
= x = 9 cm
∴ Another side,
= 9 + 8 = 17 cm
Therefore, the two parallel sides are 9 cm and 17 cm.
Question no – (13)
Solution :
Let, common factor = x
∴ Parallel side ratio = 3 : 5
∴ One side = 3x
∴ Another side = 5x
Now, according to questions,
= 1/2 × (3x + 5x) × 12 = 384
or, 8x = 384 × 2/12 = 64
or, x = 64/8 = 8 cm
∴ One side = 8 × 3 = 24 cm
∴ Another side = 8 × 5 = 40 cm
Therefore, the length parallel sides are 24 cm and 40 cm.
Question no – (14)
Solution :
Let, One side = x
Another side = 2x
Now, as per the question,
= 1/2 × (x + 2x) × 100 = 10500
or, 3x + 100/2 = 1050
or, x = 10500 × 2/100 × 3
∴ x = 70 m
∴ One side = 70 m
∴ Another side = 2 × 70 = 140 m
Therefore, the length of the side along the river are 70 m and 140 m.
Question no – (15)
Solution :
According to the given question,
The area of a trapezium is = 1586 cm2
Distance between the parallel sides is = 26 cm.
One of the parallel sides is = 38 cm
Let, the another side = x
Now, according to the question,
= 1/2 × (38 + x) × 26 = 1586
or, 38 + x = 1586 × 2/26 = 122
or, x = 122 – 38
∴ x = 84 cm
Therefore, other parallel side will be 84 cm.
Question no – (16)
Solution :
Let, PD = QC = h
AP = QB = 25 – 13/2 = 12/2 = 6 cm
∴ h2 = (10)2 – (6)2 = 100 – 36 = 64
∴ or, h = √64
= h = 8
∴ Area of ABCD,
= 1/2 × (13 + 25) × 8
= 1/2 × 38 × 8
= 152 cm2
Therefore, the area of the trapezium will be 152 cm2
Question no – (17)
Solution :
Let, AP = BQ = h
∴ DP = QC = 25 – 13/2 = 12/2 = 6 cm
= h2 = (15)2 – 62 = 225 – 36 = 189
or, h = √189
or, h = 57 √21 cm2
Therefore, the area of a trapezium will be 57 √21 cm2
Question no – (18)
Solution :
As per the given question,
Area of a trapezium is = 28 cm2
one parallel sides is = 6 cm
Altitude is = 4 cm
Let, other side is = x cm
Now, according to questions,
= 1/2 × (x + 6) × 4 = 28
or, x + 6 = 28 × 2/4 = 14
or, x = 14 – 6
∴ x = 8 cm
Therefore, the other parallel side will be 8 cm.
Question no – (19)
Solution :
Let, CO = h cm
BE = 22 – 10 = 12 cm
∴ OE = 12/2 6 cm
∴ h2 = 102 – 62 = 100 – 36
or, h = √ 64 = 8 cm
∴ Area of ABCD,
= 1/2 × (10 + 22) × 8
= 32 × 4
= 128 cm2
Thus, the area of the trapezium will be 128 cm2
Question no – (20)
Solution :
∴ Area of ·AGFO = 4 × 4 = 16 cm2
OE = 4 + 4 = 8 cm
∴ Area of · EOQD
= 8 × 4 = 32 cm2
∴ QB = 12 – (4 + 4)
= 12 – 8
= 4 cm
Area of QDCB = 1/2 × (8 + 3) × 42
= 11 × 2
= 22 cm2
∴ Total area,
= 16 + 32 + 22
= 70 cm2
Therefore, the area of the field will be 70 cm2
Mensuration – I Area of a trapezium and a polygon Exercise 20.3 Solution :
Question no – (1)
Solution :
Given in the question,
AD = 10 cm
AG = 8 cm
AH = 6 cm
AP = 5 cm
BF = 5 cm
CG = 7 cm
EH = 3 cm
AG – AF = GF = 8 – 5 = 3 cm
GD = AD – AG = 10 – 8 = 2 cm
HD = AD – AH = 10 – 6 = 4 cm
∴ Area of pentagon = area of [△ABF + BFGC + △CGD + △ADE]
= [1/2 × AF × BF] + [1/2 × (CG + BF) × GF] + [1/2 CG × GD] + [1/2 × AD × EH]
= [1/2 × 5 × 5] + [1/2 × (7 + 5) × 3] + [1/2 × 7 × 2] + [1/2 × 10 × 3]
= 25/2 + [1/2 × 12 × 3] + 7 + 15
= 25/2 + 18 + 7 + 15
= 52.5 cm²
Therefore, the area of the pentagon will be 52.5 cm²
Question no – (2)
Solution :
Figure – (i)
Given, FC = ED = 18 cm
∴ Area of ABCDEF,
= Area of [ABCF + EDCF]
= [1/2 × h × (AB + FC)] + [18 × 18]
= [1/2 × 8 × (7 + 18)] + [324]
= 4 × (25) + 324
= 100 + 324
= 424 cm²
Figure – (ii)
FC = ED = 15 cm
= h = 28 – 20 cm
= h = 8 cm
∴ Area of ABCDEF = area of [ABCD + CDEF]
= [1/2 × H × (AB + FC)] + [15 × 20]
= [1/2 × 8 × (6 + 15)] + 300
= (4 × 21) + 300 = 84 + 300
= 384 cm²
Figure – (iii)
AM = BM
= √52 – 42
= 25 – 16
= √9
= 3
∴ Area of ABQNCDMP,
= Area of [△APM + ABCD + △BQN]
= [1/2 × 3 × 4] + [AD × AB] + [1/2 × 3 × 4]
= 6 + 6 + [(3 + 4) × 6]
= 12 + (7 × 6)
= 12 + 42
= 54 cm²
Question no – (3)
Solution :
Height = 30 – 15
= 15 m
Area of park,
= area of [△ABE + BCD]
= [1/2 × 15 × 15) + 225 = (1/2 × 225) + 225
= 112.5 + 225
= 337.5 m2
Therefore, the area of this park is 337.5 m2
Question no – (4)
Solution :
AL = 10 cm FM = 20 cm
AM = 20 cm EO = 60 cm
AN = 50 cm LB = 30 cm
AO = 60 cm OC = 40 cm
AD = 90 cm.
MO = AO – AM = 60 – 20 = 40 cm
DO = AD – AO = 90 – 60 = 30 cm
LO = AO = AL = 60 – 10 = 50 cm
ND = AD – AN = 90 – 50 = 40 cm
∴ Area of polygon,
= area of [△ AFM + EFMO + △EOD + △ABL + LBCO + △COD]
= [1/2 × AM × FM] + [1/2 × LO × (FM + EO)] + [1/2 × EO + OD] + [1/2 × AL × LB] + [1/2
× LO × (LB + NC] + [1/2 × OC × OD]
= [1/2 × 20 × 20] + [1/2 × 50 × (20 + 60)] + [1/2 × 60 × 30] + [1/2 × 10 × 30] + [1/2 × 50
× (30 + 40)] + [1/2 × 40 × 30]
= 5050 cm²
Therefore, the area of the polygon will be 5050 cm²
Next Chapter Solution :
👉 Chapter 21 👈
