# 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, = d= 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

dof 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 :

Updated: June 14, 2023 — 7:00 am