Rs Aggarwal 2017 Solutions for Class 8 Math Chapter 18 Area Of A Trapezium And A Polygon are provided here with simple step-by-step explanations. These solutions for Area Of A Trapezium And A Polygon are extremely popular among Class 8 students for Math Area Of A Trapezium And A Polygon Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 8 Math Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 207:

Question 1:

Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.

Answer:

Area of a trapezium=12×Sum of parallel sides×Distance between them

                      =12×24+20×15 cm2=12×44×15 cm2=22×15 cm2=330 cm2

Hence, the area of the trapezium is 330 cm2.

Page No 207:

Question 2:

Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and the distance between them is 16 cm.

Answer:

Area of a trapezium=12×Sum of parallel sides×Distance between them

                      =12×38.7+22.3×16 cm2=12×61×16 cm2=61×8 cm2=488 cm2

Hence, the area of the trapezium is 488 cm2.

Page No 207:

Question 3:

The shape of the top surface of a table is trapezium. Its parallel sides are 1 m and 1.4 m and the perpendicular distance between them is 0.9 m. Find its area.

Answer:

Area of a trapezium=12×Sum of parallel sides×Distance between them
                       =12×1+1.4×0.9 m2=12×2.4×0.9 m2=1.2×0.9 m2=1.08 m2
Hence, the area of the top surface of the table is 1.08 m2.

Page No 207:

Question 4:

The area of a trapezium is 1080 cm2. If the lengths of its parallel sides be 55 cm and 35 cm, find the distance between them.

Answer:

Let the distance between the parallel sides be x. Now,Area of trapezium=12×55+35×x cm2

                    =12×90×xcm2=45x cm2
Area of the trapezium=1080 cm2 Given45x=1080x=108045x=24 cmHence, the distance between the parallel sides is 24 cm.

Page No 207:

Question 5:

A field is in the form of a trapezium. Its area is 1586 m2 and the distance between its parallel sides is 26 m. If one of the parallel sides is 84 m, find the other.

Answer:

Let the length of the required side be x cm.Now,Area of trapezium=12×84+x×26 m2

                        =1092+13x m2

Area of trapezium=1586 m2  Given1092+13x=158613x=1586-109213x=494x=49413x=38 mHence, the length of the other side is 38 m.

Page No 207:

Question 6:

The area of a trapezium is 405 cm2. Its parallel sides are in the ratio 4 : 5 and the distance between them is 18 cm. Find the length of each of the parallel sides.

Answer:

Let the lengths of the parallel sides of the trapezium be 4x cm and 5x cm, respectively. Now,Area of trapezium=12×4x+5x×18 cm2 
                                =12×9x×18cm2=81x cm2
Area of trapezium=405 cm2  Given81x=405x=40581x=5 cmLength of one side=4×5 cm=20 cm Length of the other side=5×5 cm=25 cm

Page No 207:

Question 7:

The area of a trapezium is 180 cm2 and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides.

Answer:

Let the lengths of the parallel sides be x cm and x+6 cm.Now, Area of trapezium=12×x+x+6×9 cm2
                     =12×2x+6×9 cm2=4.52x+6 cm2=9x+27 cm2

Area of trapezium=180 cm2 Given9x+27=1809x=180-279x=153x=1539x=17Hence, the lengths of the parallel sides are 17 cm and 23 cm, that is, 17+6 cm. 

Page No 207:

Question 8:

In a trapezium-shaped field, one of the parallel sides is twice the other. If the area of the field is 9450 m2 and the perpendicular distance between the two parallel sides is 84 m, find the length of the longer of the parallel sides.

Answer:

Let the lengths of the parallel sides be x cm and 2x cm. Area of trapezium=12×x+2x×84 m2
                    =12×3x×84 m2=42×3x m2=126x m2

Area of the trapezium=9450 m2 Given126x=9450x=9450126x=75Thus, the length of the parallel sides are 75 m and 150 m, that is, 2×75 m, and the length of the longer side is 150 m.

Page No 207:

Question 9:

The length of the fence of a trapezium-shaped field ABCD is 130 m and side AB is perpendicular to each of the parallel sides AD and BC. If BC = 54 m, CD = 19 m and AD = 42 m, find the area of the field.

Answer:

Length of the side AB=130-54+19+42 m
                         =15 m
Area of the trapezium-shaped field=12×AD+BC×AB
                                        =12×42+54×15 m2=12×96×15 m2=48×15 m2=720 m2

Hence, the area of the field is 720 m2.

Page No 207:

Question 10:

In the given figure, ABCD is a trapezium in which AD||BC, ∠ABC = 90°, AD = 16 cm, AC = 41 cm and BC = 40 cm. Find the area of the trapezium.

Answer:

ABC=90°From the right ABC, we have:AB2=AC2-BC2AB2=412-402AB2=1681-1600AB2=81AB=81AB=9 cm Length AB=9 cmNow,Area of the trapezium=12×AD+BC×AB
                         =12×16+40×9 cm2=12×56×9 cm2=28×9 cm2=252 cm2
Hence, the area of the trapezium is 252 cm2.

Page No 207:

Question 11:

The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides are both equal, each being 13 cm. Find the area of the trapezium.

Answer:


Let ABCD be the given trapezium in which ABDC, AB=20 cm, DC=10 cm and AD=BC=13 cm.Draw CLAB and CMDA meeting AB at L and M, respectively.Clearly, AMCD is a parallelogram.Now,AM=DC=10 cmMB=AB-AM     =20-10 cm     =10 cmAlso, CM=DA=13 cmTherefore, CMB is an isosceles triangle and CLMB.L is the midpoint of B.ML=LB=12×MB 
              =12×10 cm=5 cm

From right CLM, we have:CL2=CM2-ML2 cm2CL2=132-52 cm2CL2=109-25 cm2CL2=144 cm2CL=144 cmCL=12 cm Length of CL=12 cmArea of the trapezium=12×AB+DC×CL
                         =12×20+10×12 cm2=12×30×12 cm2=15×12 cm2=180 cm2
Hence, the area of the trapezium is 180 cm2.

Page No 207:

Question 12:

The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. Find the area of the trapezium.

Answer:



Let ABCD be the trapezium in which ABDC, AB=25 cm, CD=11 cm, AD=13 cm and BC=15 cm.Draw CLAB and CMDA meeting AB at L and M, respectively.Clearly, AMCD is a parallelogram.Now,MC=AD=13 cmAM=DC=11 cmMB=AB-AM
          =25-11 cm=14 cm

Thus, in CMB, we have:CM=13 cmMB=14 cm BC=15 cm s=1213+14+15 cm       =1242 cm       =21 cms-a=21-13 cm          =8 cms-b=21-14 cm          =7 cms-c=21-15cm          =6 cm Area of CMB=ss-as-bs-c 
                      =21×8×7×6  cm2=84  cm2

 12×MB×CL=84 cm212×14×CL=84 cm2
 CL=847CL=12 cm
Area of the trapezium=12×AB+DC×CL
                         =12×25+11×12 cm2=12×36×12 cm2=18×12 cm2=216 cm2

Hence, the area of the trapezium is 216 cm2.



Page No 210:

Question 1:

In the given figure, ABCD is a quadrilateral in which AC = 24 cm, BL ⊥ AC and DM ⊥ AC such that BL = 8 cm and DM = 7 cm. Find the area of quad. ABCD.

Answer:

Area of quadrilateral ABCD=Area of ADC+Area of ACB
                                             =12×AC×DM+12×AC×BL=12×24×7+12×24×8 cm2=84+96 cm2=180 cm2

Hence, the area of the quadrilateral is 180 cm2.

Page No 210:

Question 2:

In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD is 36 m, ALBD and CMBD such that AL = 19 m and CM = 11 m. Find the area of the field.

Answer:

Area of quadrilateral ABCD=Area of ABD+Area of BCD
                                         =12×BD×AL+12×BD×CM
                                         =12×36×19+12×36×11 m2=342+198 m2=540 m2

Hence, the area of the field is 540 m2.

Page No 210:

Question 3:

Find the area of pentagon ABCDE in which BLAC, DMAC and ENAC such that AC = 18 cm, AM = 14 cm, AN = 6 cm, BL = 4 cm, DM = 12 cm and EN = 9 cm.

Answer:

Area of pentagon ABCDE=Area of AEN+Area of trapezium EDMN+Area of DMC+Area of ACB
                                      =12×AN×EN+12×EN+DM×NM+12×MC×DM+12×AC×BL=12×AN×EN+12×EN+DM×AM-AN+12×AC-AM×DM+12×AC×BL=12×6×9+12×9+12×14-6+12×18-14×12+12×18×4 cm2=27+84+24+36 cm2=171 cm2

Hence, the area of the given pentagon is 171 cm2.

Page No 210:

Question 4:

Find the area of hexagon ABCDEF in which BLAD, CMAD, ENAD and FPAD such that AP = 6 cm, PL = 2 cm, LN = 8 cm, NM = 2 cm, MD = 3 cm, FP = 8 cm, EN = 12 cm, BL = 8 cm and CM = 6 cm.

Answer:

Area of hexagon ABCDEF=Area ofAFP+Area of trapezium FENP+Area ofALB
=12×AP×FP+12×FP+EN×PN+12×ND×EN+12×MD×CM+12×CM+BL×LM+12×AL×BL=12×AP×FP+12×FP+EN×PL+LN+12×NM+MD×CM+12×MD×CM+12×CM+BL×LN+NM+12×AP+PL×BL=12×6×8+12×8+12×2+8+12×2+3×12+12×3×6+12×6+8×8+2+12×6+2×8 cm2=24+100+30+9+70+32 cm2=265 cm2

Hence, the area of the hexagon is 265 cm2.

Page No 210:

Question 5:

Find the area of pentagon ABCDE in which BLAC, CMAD and ENAD such that AC = 10 cm, AD = 12 cm, BL = 3 cm, CM = 7 cm and EN = 5 cm.

Answer:

Area  of pentagon ABCDE=Area of ABC+Area of ACD+Area of ADE
                                             =12×AC×BL+12×AD×CM+12×AD×EM=12×10×3+12×12×7+12×12×5 cm2=15+42+30 cm2=87 cm2

Hence, the area of the pentagon is 87 cm2.

Page No 210:

Question 6:

Find the area enclosed by the given figure ABCDEF as per dimensions given herewith.

Answer:

Area enclosed by the given figure=Area of trapezium FEDC+Area of square ABCF
                                                 =12×6+20×8+20×20cm2=104+400cm2=504 cm2

Hence, the area enclosed by the figure is 504 cm2.



Page No 211:

Question 7:

Find the area of given figure ABCDEFGH as per dimensions given in it.

Answer:

We will find the length of AC.From the right triangles ABC and HGF, we have:AC2=HF2=52-42 cm
                =25-16cm=9 cm

AC=HF=9 cm               =3 cm
Area of the given figure ABCDEFGH=Area of rectangle ADEH+Area of ABC+Area of HGF
                                                      =Area of rectangle ADEH+2Area of ABC=AD×DE+2Area of ABC=AC+CD×DE+212×BC×AC=3+4×8+212×4×3 cm2=56+12 cm=68 cm2
Hence, the area of the given figure is 68 cm2.

Page No 211:

Question 8:

Find the area of a regular hexagon ABCDEF in which each side measures 13 cm and whose height is 23 cm, as shown in the given figure.

Answer:

Let AL=DM=x cm LM=BC=13 cm x+13+x=232x+13=232x=23-132x=10x=5 AL=5 cmFrom the right AFL, we have:    FL2=AF2-AL2FL2=132-52FL2=169-25FL2=144FL=144FL=12 cm FL=BL=12 cmArea of a regular hexagon=Area of the trapezium ADEF+Area of the trapezium ABCD
                                  =2Area of trapezium ADEF=212×AD+EF×FL=212×23+13×12cm2=212×36×12cm2=432 cm2
Hence, the area of the given regular hexagon is 432 cm2.

Page No 211:

Question 1:

Tick (✓) the correct answer:
The parallel sides of a trapezium measure 14 cm and 18 cm and the distance between them is 9 cm. The area of the trapezium is
(a) 96 cm2
(b) 144 cm2
(c) 189 cm2
(d) 207 cm2

Answer:

(b) 144 cm2
Area of the trapezium=12×14+18×9 cm2
                            =12×32×9 cm2=144 cm2

Page No 211:

Question 2:

Tick (✓) the correct answer:
The lengths of the parallel sides of a trapezium are 19 cm and 13 cm and its area is 128 cm2. The distance between the parallel sides is
(a) 9 cm
(b) 7 cm
(c) 8 cm
(d) 12.5 cm

Answer:

(c) 8 cm

Let the distance between the parallel sides be x cm.Then, area of the trapezium=12×19+13×x cm2
                                    =12×32×x cm2=16x cm2
But it is given that the area of the trapezium is 128 cm2.
 16x=128x=12816x=8 cm

Page No 211:

Question 3:

Tick (✓) the correct answer:
The parallel sides of a trapezium are in the ratio 3 : 4 and the perpendicular distance between them is 12 cm. If the area of the trapezium is 630 cm2, then its shorter of the parallel sides is
(a) 45 cm
(b) 42 cm
(c) 60 cm
(d) 36 cm

Answer:

(a) 45 cm

Let the length of the parallel sides be 3x cm and 4x cm, respectively.Then, area of the trapezium=12×3x+4x×12 cm2
                                    =12×7x×12cm2=42 x cm2

But it is given that the area of the trapezium is 630 cm2.

 42x=630x=63042x=15 cm
Length of the parallel sides = 3×15cm=45 cm                                        4×15 cm=60 cmHence, the shorter of the parallel sides is 45 cm.

Page No 211:

Question 4:

Tick (✓) the correct answer:
The area of a trapezium is 180 cm2 and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, the length of the longer of the parallel sides is
(a) 17 cm
(b) 23 cm
(c) 18 cm
(d) 24 cm

Answer:

(b) 23 cm

Let the length of the parallel sides be x cm and x+6 cm, respectively.Then, area of the trapezium=12×x+x+6×9 cm2
                                    =12×2x+6×9 cm2=4.52x+6 cm2=9x+27 cm2
But it is given that the area of the trapezium is 180 cm2. 9x+27=1809x=180-279x=153x=1539x=17Therefore, the length of the parallel sides are 17 cm and 17+6 cm, which is equal to 23 cm.Hence, the length of the longer parallel side is 23 cm.

Page No 211:

Question 5:

Tick (✓) the correct answer:
In the given figure, AB||DC and DA ⊥ AB. If DC = 7 cm, BC = 10 cm, AB = 13 cm and CL ⊥ AB, the area of trap. ABCD is
(a) 84 cm2
(b) 72 cm2
(c) 80 cm2
(d) 91 cm2

Answer:

(c) 80 cm2

From the given trapezium, we find:DC=AL=7 cm     since DAAB and CLABFrom the right CBL, we have:    CL2=CB2-LB2CL2=102-62CL2=100-36CL2=64CL=64CL=8 cmArea of the trapezium=12×7+13×8 cm2
                            =12×20×8 cm2=80 cm2



Page No 212:

Question 1:

The base of a triangular field is three times its height and its area is 1350 m2. Find the base and height of the field.

Answer:

Let the base of the triangular field be 3x cm and its height be x cm.Then, area of the triangle=12×3x×x m2
                                 =3x22 m2
But it is given that the area of the triangular field is 1350 m2.
 3x22=1350x2=1350×23x2=900x=900x=30 m
Hence, the height of the field is 30 m.Its base = 3×30 m=90 m

Page No 212:

Question 2:

Find the area of an equilateral triangle of side 6 cm.

Answer:

Area of an equilateral traingle=34×side2 square units                                            = 34×6×6 cm2
                                       =34×36 cm2=93 cm2
Hence, the area of an equilateral triangle is 93 cm2.

Page No 212:

Question 3:

The perimeter of a rhombus is 180 cm and one of its diagonals is 72 cm. Find the length of the other diagonal and the area of the rhombus.

Answer:



Let ABCD be a rhombus whose diagonals AC and BD intersect at a point O.Let the length of the diagonal AC be 72 cm and the side of the rhombus be x cm.Perimeter of the rhombus=4x cmBut it is given that the perimeter of the rhombus is180 cm. 4x=180 x=1804x=45Hence, the length of the side of the rhombus is 45 cm.We know that the diagonals of the rhombus bisect each other at right angles. AO=12ACAO=12×72cmAO=36 cmFrom right AOB, we have:    BO2 =AB2-AO2BO2=452-362BO2=2025-1296BO2=729BO=729BO=27 cm BD=2×BOBD=2×27cmBD=54 cmHence, the length of the other diagonal is 54 cm.Area of the rhombus=12×72×54  cm2
=1944 cm2

Page No 212:

Question 4:

The area of a trapezium is 216 m2 and its height is 12 m. If one of the parallel sides is 14 m less than the other, find the length of each of the parallel sides.

Answer:

Let the length of the parallel sides be x m and x-14 m.Then, area of the trapezium=12×x+x-14×12 m2
                                     =62x-14m2=12x-84m2
But it is given that the area of the trapezium is 216 m2. 12x-84=216 12x=216+84 12x=300 x=30012 x=25Hence, the length of the parallel sides are 25 m and 25-14m, which is equal to 11 m.

Page No 212:

Question 5:

Find the area of a quadrilateral one of whose diagonals is 40 cm and the lengths of the perpendiculars drawn from the opposite vertices on the diagonal are 16 cm and 12 cm.

Answer:


Let ABCD be a quadilateral.Diagonal, AC=40 cmBLAC, such that BL=16 cmDMAC, such that DM=12 cmArea of the quadilateral=Area of DAC+Area of ACB

                                       =12×AC×DM+12×AC×BL cm2=12×40×12+12×40×16 cm2=240+320 cm2=560 cm2

Hence, the area of the quadilateral is 560 cm2.

Page No 212:

Question 6:

A field is in the form of a right triangle with hypotenuse 50 m and one side 30 m. Find the area of the field.

Answer:

Let the other side of the triangular field be  x m.    x2=502-302x2=2500-900x2=1600x=1600x=40 Area of the field=12×30×40 m2
                          
                          =600 m2

Page No 212:

Question 7:

Mark (✓) against the correct answer:
The base of a triangle is 14 cm and its height is 8 cm. The area of the triangle is
(a) 112 cm2
(b) 56 cm2
(c) 122 cm2
(d) 66 cm2

Answer:

(b) 56 cm2
Area of the triangle=12×14×8 cm2
                          =56 cm2

Page No 212:

Question 8:

Mark (✓) against the correct answer:
The base of a triangle is four times its height and its area is 50 m2. The length of its base is
(a) 10 m
(b) 15 m
(c) 20 m
(d) 25 m

Answer:

(c) 20 m

Let the height of the triangle be x m and its base be 4x m respectively.Then, area of the triangle=12×4x×x m2
                                 =2x2 m2
But, the area of the triangle is 50 m2. 2x2=50x2=502x2=25x=25x=5Length of its base = 4×5 m=20 m

Page No 212:

Question 9:

Mark (✓) against the correct answer:
The diagonal of a quadrilateral is 20 cm in length and the lengths of perpendiculars on it from the opposite vertices are 8.5 cm and 11.5 cm. The area of the quadrilateral is
(a) 400 cm2
(b) 200 cm2
(c) 300 cm2
(d) 240 cm2

Answer:

(b) 200 cm2


Let ABCD be a quadilateral. Diagonal, AC=20 cmBLAC, such that BL=8.5 cmDMAC, such that DM=11.5 cmArea of the quadilateral=Area of DAC+Area of ACB
                                       =12×AC×DM+12×AC×BL cm2=12×20×11.5+12×20×8.5 cm2=85+115 cm2=200 cm2

Page No 212:

Question 10:

Mark (✓) against the correct answer:
Each side of a rhombus is 15 cm and the length of one of its diagonals is 24 cm. The area of the rhombus is
(a) 432 cm2
(b) 216 cm2
(c) 180 cm2
(d) 144 cm2

Answer:

(b) 216 cm2


Let ABCD be a rhombus whose diagonals AC and BD intersect at a point O.Let the length of the diagonal AC be 24 cm and the side of the rhombus be 15 cm.We know that the diagonals of the rhombus bisect each other at right angles.   AO=12ACAO=12×24 cmAO=12 cmFrom right AOB, we have:    BO2 =AB2-AO2BO2=152-122BO2=225-144BO2=81BO=81BO=9 cmBD=2×BOBD=2×9 cmBD=18 cmHence, the length of the other diagonal is 18 cm.Area of the rhombus=12×24×18 cm2

                           =216 cm2

Page No 212:

Question 11:

Mark (✓) against the correct answer:
The area of a rhombus is 120 cm2 and one of its diagonals is 24 cm. Each side of the rhombus is
(a) 10 cm
(b) 13 cm
(c) 12 cm
(d) 15 cm

Answer:

(b) 13 cm



Let ABCD be a rhombus whose diagonals AC and BD intersect at a point O.Let the length of the diagonal AC be 24 cm.Area of the rhombus=12×AC×BD cm2But the area of the rohmbus is 120 cm2   (given)12×AC×BD = 120or 12×24×BD = 120or 12×BD = 120or BD = 12012 = 10 cm OB = BD2 = 102 = 5 cm

And OA = AC2 = 242 = 12 cmNow, in right triangle AOB:  AB2 = OA2 + OB2or AB2 = 122 + 52                = 144 + 25                = 169or AB = 169 = 13 cmTherefore, each side of the rhombus is 13 cm.                               

Page No 212:

Question 12:

Mark (✓) against the correct answer:
The parallel sides of a trapezium are 54 cm and 26 cm and the distance between them is 15 cm. The area of the trapezium is
(a) 702 cm2
(b) 810 cm2
(c) 405 cm2
(d) 600 cm2

Answer:

(d) 600 cm2

Area of the trapezium=12×54+26×15 cm2
                            =12×80×15 cm2
                             =
600 cm2

Page No 212:

Question 13:

Mark (✓) against the correct answer:
The area of a trapezium is 384 cm2. Its parallel sides are in the ratio 5 : 3 and the distance between them is 12 cm. The longer of the parallel sides is
(a) 24 cm
(b) 40 cm
(c) 32 cm
(d) 36 cm

Answer:

(b) 40 cm

Let the length of the parallel sides be 5x cm and 3x cm, respectively.Area of the trapezium=12×5x+3x×12 cm2
                                    =12×8x×12 cm2=48x cm2

But, the area of the trapezium is 384 cm2.48x = 384x=38448=8Longer side = 5x=5×8=40 cm

Page No 212:

Question 14:

Fill in the blanks.
(i) Area of triangle = 12×(.........)×(.........).
(ii) Area of a ||gm = 12×(.........)×(.........).
(iii) Area of a trapezium = 12×(.........)×(.........).
(iv) The parallel sides of a trapezium are 14 cm and 18 cm and the distance between them is 8 cm. The area of the trapezium is ......... cm2.

Answer:

i Area of a triangle=12×Base×Heightii Area of a gm=Base×Heightiii Area of a trapezium=12×Sum of the parallel sides×Distance between themiv Area of a trapezium=12×14+18×8 cm2
                               =12×32×8cm2=128 cm2



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