Rd Sharma 2018 Solutions for Class 8 Math Chapter 20 Mensuration I Area Of A Trapezium And A Polygon are provided here with simple step-by-step explanations. These solutions for Mensuration I Area Of A Trapezium And A Polygon are extremely popular among Class 8 students for Math Mensuration I 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 Rd Sharma 2018 Book of Class 8 Math Chapter 20 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2018 Solutions. All Rd Sharma 2018 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 20.13:

Question 1:

A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2?

Answer:

Given:Base of a flooring tile that is in the shape of a parallelogram = b=24 cmCorresponding height = h=10 cmNow,in a parallelogram:Area(A)= Base (b)×Height (h) Area of a tile= 24 cm×10 cm=240 cm2Now, observe that the area of the floor is 1080 m2.1080 m2=1080×1m×1m=1080×100 cm×100 cm                     ( Because 1 m= 100 cm)=1080×100×100×cm×cm=10800000 cm2 Number of required tiles=10800000240=45000Hence, we need 45000 tiles to cover the floor.

Page No 20.13:

Question 2:

A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in Fig. 20.23. If AB = 60 m and BC = 28 m, find the area of the plot.

Answer:

The given figure has a rectangle with a semicircle on one of its sides.   



Total area of the plot=Area of rectangle ABCD + Area of semicircle with radius (r = 282=14m) Area of the rectangular plot with sides 60m and 28m =60×28=1680 m2         ...(i)And, area of the semicircle with radius 14m =12π×(14)2=12×227×14×14=308m2    ...(ii) Total area of the plot =1680+308=1988m2           ...(from (i) and (ii))

Page No 20.13:

Question 3:

A playground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π = 22/7).

Answer:

It is given that the playground is in the shape of a rectangle with two semicircles on its smaller sides. Length of the rectangular portion is 36 m and its width is 24.5 m as shown in the figure below.


Thus, the area of the playground will be the sum of the area of a rectangle and the areas of the two semicircles with equal diameter 24.5 m.Now, area of rectangle with length 36m and width 24.5m:Area of rectangle=length×width=36m×24.5 m =882 m2Radius of the semicircle = r = diameter2=24.52=12.25m Area of the semicircle=12πr2=12×227×(12.25)2=235.8 m2 Area of the complete playground= area of the rectangular ground+ 2×area of a semicircle=882+2×235.8=1353.6 m2

Page No 20.13:

Question 4:

A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 m have been cut. Find the area of the remaining part.

Answer:

It is given that the length of the rectangular piece is 20 m and its width is 15 m.And, from each corner a quadrant each of radius 3.5 m has been cut out.A rough figure for this is given below:



 Area of the remaining part=Area of the rectangular piece-(4×Area of a quadrant of radius 3.5m)Now, area of the rectangular piece=20×15=300 m2And, area of a quadrant with radius 3.5 m =14πr2=14×227×(3.5)2=14×227×3.5×3.5=9.625 m2 Area of the remaining part=300-(4×9.625)=261.5 m2



Page No 20.14:

Question 5:

The inside perimeter of a running track (shown in Fig. 20.24) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track.

Answer:



It is given that the inside perimeter of the running track is 400 m. It means the length of the inner track is 400 m.Let r be the radius of the inner semicircles.Observe: Perimeter of the inner track=Length of two straight portions of 90 m+Length of two semicircles 400=(2×90)+(2×Perimiter of a semicircle)400 = 180+(2×227×r)400-180=(447×r)447×r=220r=220×744=35 m Width of the inner track=2r=2×35=70 mSince the track is 14 m wide at all places, so the width of the outer track: 70+(2×14)=98 m Radius of the outer track semicircles=982=49 mArea of the outer track=(Area of the rectangular portion with sides 90 m and 98 m)+ (2×Area of two semicircles with radius 49 m)=(98×90)+(2×12×227×492)=(8820)+(7546)=16366 m2And, area of the inner track=(Area of the rectangular portion with sides 90 m and 70 m)+ (2×Area of the semicircle with radius 35 m)=(70×90)+(2×12×227×352)=(6300)+(3850)=10150 m2 Area of the running track=Area of the outer track-Area of the inner track=16366-10150=6216 m2And, length of the outer track=(2×length of the straight portion)+(2×perimeter of the semicircles with radius 49 m)=(2×90)+(2×227×49)=180+308=488 m

Page No 20.14:

Question 6:

Find the area of Fig. 20.25, in square cm, correct to one place of decimal. (Take π = 22/7)

Answer:

The given figure is:  



Construction : Connect A to D.Then, we have: Area of the given figure =  (Area of rectangle ABCD + Area of the  semicircle) - (Area of  AED). Total area of the figure=(Area of rectangle with sides 10 cm and 10 cm)+(Area of semicircle with radius=102=5 cm)-(Area of triangle AED with base 6 cm and height 8 cm)=(10×10)+(12×227×52)-(12×6×8)=100+39.3-24=115.3 cm2

Page No 20.14:

Question 7:

The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in kilometres per hour. [Use π = 22/7]

Answer:

It is given that the diameter of the wheel is 90 cm. Radius of the circular wheel,  r=902=45 cm. Perimeter of the wheel =2×π×r=2×227×45=282.857 cmIt means the wheel travels 282.857 cm in a revolution.Now, it makes 315 revolutions per minute. Distance travelled by the wheel in one minute=315×282.857=89100 cm Speed =89100 cm per minute=89100 cm1 minuteNow, we need to convert it into kilometers per hour. 89100 cm1 minute =89100×1100000kilometer160hour=89100100000×601×kilometerhour=53.46 kilometers per hour

Page No 20.14:

Question 8:

The area of a rhombus is 240 cm2 and one of the diagonal is 16 cm. Find another diagonal.

Answer:

Given:Area of the rhombus=240 cmLength of one of its diagonals=16 cmWe know that if the diagonals of a rhombus are d1 and d2, then the area of the rhombus is given by:Area=12(d1×d2)Putting the given values:240=12(16×d2)240×2=16×d2This can be written as follows:16×d2=480d2=48016d2=30 cmThus, the length of the other diagonal of the rhombus is 30 cm.

Page No 20.14:

Question 9:

The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Answer:

Given:Lengths of the diagonals of a rhombus are 7.5 cm and 12 cm.Now, we know: Area=12(d1×d2)  Area of rhombus=12(7.5×12)=45 cm2

Page No 20.14:

Question 10:

The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

Answer:




Given: Diagonal of a quadrilateral shaped field=24 mPerpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m.Now, we know:Area=12×d×(h1+h2) Area of the field=12×24×(8+13)=12×21=252 m2

Page No 20.14:

Question 11:

Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.

Answer:

Given: Side of the rhombus=6 cmAltitude=4 cm One of the diagonals = 8 cmArea of the rhombus =  Side × Altitude = 6×4 = 24 cm2     ........(i)We know: Area of rhombus = 12×d1×d2 Using (i):24 =  12×d1×d2 24 = 12×8×d2d2 = 6 cm

Page No 20.14:

Question 12:

The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is Rs 4.

Answer:

Given:The floor consist of 3000 rhombus shaped tiles.The lengths of the diagonals of each tile are 45 cm and 30 cm. Area of a rhombus shaped tile =12×(45×30)=675 cm2 Area of the complete floor=3000×675=2025000 cm2Now, we need to convert this area into m2 because the rate of polishing is given as per m2. 2025000 cm2=2025000×cm×cm=2025000×1100 m×1100 m=202.5 m2Now, the cost of polishing 1 m2 is Rs 4. Total cost of polishing the complete floor=202.5×4=810Thus, the total cost of polishing the floor is Rs 810.

Page No 20.14:

Question 13:

A rectangular grassy plot is 112 m long and 78 m broad. It has a gravel path 2.5 m wide all around it on the side. Find the area of the path and the cost of constructing it at Rs 4.50 per square metre.

Answer:

Given:The length of a rectangular grassy plot is 112 m and its width is 78 m.Also, it has a gravel path of width 2.5 m around it on the sides .Its rough diagram is given below:



Length of the inner rectangular field=112-(2×2.5)=107 mThe width of the inner rectangular field=78-(2×2.5)=73 m Area of the path=(Area of the rectangle with sides 112 m and 78 m)-(Area of the rectangle with sides 107 m and 73 m)=(112×78)-(107×73)=8736-7811=925 m2Now, the cost of constructing the path is Rs 4.50 per square meter. Cost of constructing the complete path=925×4.50=Rs 4162.5Thus, the total cost of constructing the path is Rs 4162.5

Page No 20.14:

Question 14:

Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.

Answer:

Given:Side of the rhombus=20 cm Length of a diagonal=24 cmWe know: If d1 and d2 are the lengths of the diagonals of the rhombus, thenside of the rhombus=12d12+d22So, using the given data to find the length of the other diagonal of the rhombus:20=12242+d2240=242+d22Squaring both sides to get rid of the square root sign:402=242+d22d22=1600-576=1024d2=1024=32 cm Area of the rhombus=12(24×32)=384 cm2

Page No 20.14:

Question 15:

The length of a side of a square field is 4 m. what will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonal is 2 m?

Answer:

Given:Length of the square field = 4 m∴ Area of the square field = 4×4 = 16 m2Given: Area of the rhombus = Area of the square fieldLength of one diagonal of the rhombus = 2 m∴ Side of the rhombus=12d12+d22And, area of the rhombus=12×(d1×d2)∴ Area:16=12(2×d2)d2=16 mNow, we need to find the length of the side of the rhombus. ∴ Side of the rhombus = 1222+162 12260 124×65 12×265 65 mAlso, we know: Area of the rhombus = Side × Altitude∴ 16=65×AltitudeAltitude=1665 m

Page No 20.14:

Question 16:

Find the area of the field in the form of a rhombus, if the length of each side be 14 cm and the altitude be 16 cm.

Answer:

Given:Length of each side of a field in the shape of a rhombus = 14 cmAltitude = 16 cmNow, we know: Area of the rhombus = Side×Altitude∴ Area of the field = 14×16 = 224 cm2

Page No 20.14:

Question 17:

The cost of fencing a square field at 60 paise per metre is Rs 1200. Find the cost of reaping the field at the rate of 50 paise per 100 sq. metres.

Answer:

Given:Cost of fencing 1 metre of a square field = 60 paiseAnd, the total cost of fencing the entire field = Rs 1200 = 1,20,000 paise∴ Perimeter of the square field = 12000060=2000 metresNow, perimeter of a square = 4×side For the given square field:4×Side = 2000 mSide = 20004= 500 metres∴ Area of the square field = 500×500=250000 m2Again, given: Cost of reaping per 100 m2 = 50 paise∴ Cost of reaping per 1 m2 =50100paise∴ Cost of reaping 250000 m2=50100×250000=125000 paiseThus, the total cost of reaping the complete square field is 125000 paise, i.e. Rs 1250.

Page No 20.14:

Question 18:

In exchange of a square plot one of whose sides is 84 m, a man wants to buy a rectangular plot 144 m long and of the same area as of the square plot. Find the width of the rectangular plot.

Answer:

Given:Side of the square plot = 84 mNow, the man wants to exchange it with a rectangular plot of the same area with length 144.Area of the square plot = 84×84 = 7056 m2∴ Area of the rectangular plot = Length×Width7056=144×WidthWidth=7056144=49 mHence, the width of the rectangular plot is 49 m.

Page No 20.14:

Question 19:

The area of a rhombus is 84 m2. If its perimeter is 40 m, then find its altitude.

Answer:

Given:Area of the rhombus = 84 m2 Perimeter = 40 mNow, we know: Perimeter of the rhombus = 4×Side∴ 40=4×SideSide=404=10 mAgain, we know: Area of the rhombus = Side×Altitude84=10×AltitudeAltitude = 8410 = 8.4 mHence, the altitude of the rhombus is 8.4 m.

Page No 20.14:

Question 20:

A garden is in the form of a rhombus whose side is 30 metres and the corresponding altitude is 16 m. Find the cost of levelling the garden at the rate of Rs 2 per m2.

Answer:

Given:Side of the rhombus shaped garden = 30 m Altitude = 16 mNow, area of a rhombus = side×altitude∴ Area of the given garden=30×16=480 m2Also, it is given that the rate of levelling the garden is Rs 2 per 1m2.∴ Total cost of levelling the complete garden of area 480 m2=480×2= Rs 960

Page No 20.14:

Question 21:

A field in the form of a rhombus has each side of length 64 m and altitude 16 m. What is the side of a square field which has the same area as that of a rhombus?

Answer:

Given:Each side of a rhombus shaped field = 64 mAltitude = 16 mWe know: Area of rhombus = Side×Altitude∴ Area of the field = 64×16=1024 m2Given: Area of the square field = Area of the rhombusWe know: Area of a square=(Side)2∴ 1024=(Side)2Side=1024=32 mThus, the side of the square field is 32 m.



Page No 20.15:

Question 22:

The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the length of the other diagonal.

Answer:

Given: Area of the rhombus = Area of the triangle with base 24.8 cm and altitude 16.5 cmArea of the triangle = 12×base×altitude = 12×24.8×16.5=204.6 cm2∴ Area of the rhombus = 204.6 cm2Also, length of one of the diagonals of the rhombus=22 cmWe know: Area of rhombus=12(d1×d2)204.6=12(22×d2)22×d2=409.2d2=409.222=18.6 cmHence, the length of the other diagonal of the rhombus is 18.6 cm.



Page No 20.22:

Question 1:

Find the area, in square metres, of the trapezium whose bases and altitudes are as under:
(i) bases = 12 dm and 20 dm, altitude = 10 dm
(ii) bases = 28 cm and 3 dm, altitude = 25 cm
(iii) bases = 8 m and 60 dm, altitude = 40 dm
(iv) bases = 150 cm and 30 dm, altitude = 9 dm.

Answer:

(i)Given: Bases:12 dm = 1210m = 1.2 mAnd, 20 dm=2010m=2 mAltitude = 10 dm = 1010m = 1 mArea of trapezium = 12×(Sum of the bases)×(Altitude)=12×(1.2+2) m×(1) m=1.6×m×m=1.6 m2(ii)Given: Bases:28 cm =28100m = 0.28 mAnd, 3 dm = 310m = 0.3 mAltitude = 25 cm = 25100m = 0.25 mArea of trapezium = 12×(Sum of the bases)×(Altitude)=12×(0.28+0.3) m×(0.25) m= 0.0725 m2(iii)Given:Bases:8 mAnd, 60 dm = 6010m = 6 mAltitude = 40 dm = 4010m = 4 mArea of trapezium=12×(Sum of the bases)×(Altitude)=12×(8+6) m×(4) m=28×m×m=28 m2(iv)Given:Bases:150 cm =150 100m = 1.5 mAnd, 30 dm = 3010m = 3 mAltitude = 9 dm = 910m = 0.9 mArea of trapezium=12×(Sum of the bases)×(Altitude)=12×(1.5+3) m×(0.9) m=2.025×m×m=2.025 m2

Page No 20.22:

Question 2:

Find the area of trapezium with base 15 cm and height 8 cm, if the side parallel to the given base is 9 cm long.

Answer:

Given:Lengths of the parallel sides are 15 cm and 9 cm.Height=8 cmArea of trapezium=12×(Sum of the opposite sides)×(Distance between the parallel sides)=12×(15+9)×(8)=96 cm2

Page No 20.22:

Question 3:

Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm and whose height is 12 dm.

Answer:

Given:Lengths of the parallel sides are 16 dm and 22 dm.And, height between the parallel sides is 12 dm.Area of trapezium = 12×(Sum of the parallel sides)×(Height)=12×(16+22)×(12)=228 dm2=228×dm×dm=228×110110m=2.28 m2

Page No 20.22:

Question 4:

Find the height of a trapezium, the sum of the lengths of whose bases (parallel sides) is 60 cm and whose area is 600 cm2.

Answer:

Given: Sum of the parallel sides of a trapezium = 60 cm Area of the trapezium = 600 cm2Area of trapezium=12×(Sum of the parallel sides)×(Height)On putting the values:600=12×60×(Height)600 = 30×(Height)Height = 60030 = 20 cm

Page No 20.22:

Question 5:

Find the altitude of a trapezium whose area is 65 cm2 and whose bases are 13 cm and 26 cm.

Answer:

Given:Area of the trapezium=65 cm2 The lengths of the opposite parallel sides are 13 cm and 26 cm.Area of trapezium=12×(Sum of parallel bases)×(Altitude)On putting the values: 65=12×(13+26)×(Altitude)65×2=39×AltitudeAltitude=13039=103 cm

Page No 20.22:

Question 6:

Find the sum of the lengths of the bases of a trapezium whose area is 4.2 m2 and whose height is 280 cm.

Answer:

Given:Area of the trapezium = 4.2 m2Height = 280 cm = 280100m = 2.8 mArea of trapezium =12×(Sum of the parallel bases)×(Height)4.2 = 12×(Sum of the parallel bases)×2.84.2×2=(Sum of the parallel bases)×2.8Sum of the parallel bases=8.42.8=3 m

Page No 20.22:

Question 7:

Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate this area as
(i) the sum of the areas of two triangles and one rectangle.
(ii) the difference of the area of a rectangle and the sum of the areas of two triangles.

Answer:

Given:Length of the parallel sides of a trapezium are 10 cm and 15 cm. The distance between them is 6 cm.Let us extend the smaller side and then draw perpendiculars from the ends of both sides.


(i)Area of trapezium ABCD =(Area of rectangle EFCD)+(Area of triangle AED+Area of triangle BFC)=(10×6)+[(12×AE×ED)+(12×BF×FC)]=60+[(12×AE×6)+(12×BF×6)]=60+[3AE+3 BF]=60+3×(AE+BF)Here,  AE+EF+FB = 15cmAnd  EF = 10 cm∴  AE+10+BF=15Or,  AE+BF=15-10=5 cmPutting this value in the above formula:Area of the trapezium=60+3×(5)=60+15=75 cm2

(ii)In this case, the figure will look as follows:


Area of trapezium ABCD=(Area of rectangle ABGH)-[(Area of triangle AHD)+(Area of triangle BGC)]=(15×6)-[(12×DH×6)+(12×GC×6)]=90-[3×DH+3×GC]=90-3[DH+GC]Here, HD+DC+CG=15 cm DC=10 cmHD+10+CG=15HD+GC=15-10=5 cmPutting this value in the above equation:Area of the trapezium=90-3(5)=90-15=75 cm2

Page No 20.22:

Question 8:

The area of a trapezium is 960 cm2. If the parallel sides are 34 cm and 46 cm, find the distance between them.

Answer:

Given:Area of the trapezium = 960 cm2 And the length of the parallel sides are 34 cm and 46 cm.Area of trapezium=12×(Sum of the parallel sides)×(Perpendicular distance between the parallel sides)960 = 12×(34+46)×(Height)960= 40×(Height)Height = 96040 = 24 cm



Page No 20.23:

Question 9:

Find the area of Fig. 20.35 as the sum of the areas of two trapezium and a rectangle.

Answer:

The given figure is:  



In the given figure, we have a rectangle of length 50 cm and width 10 cm, and two similar trapeziums with parallel sides as 30 cm and 10 cm at both ends.Suppose x is the perpendicular distance between the parallel sides in both the trapeziums.We have:Total length of the given figure= Length of the rectangle+ 2×Perpendicular distance between the parallel sides in both the trapeziums70=50+2×x2×x=70-50=20x=202=10 cmNow, area of the complete figure=(area of the rectangle with sides 50 cm and 10 cm)+2×(area of the trapezium with parallel sides 30 cm and 10 cm, and height 10 cm)=(50×10)+2×[12×(30+10)×(10)]=500+2×[200]=900 cm2

Page No 20.23:

Question 10:

Top surface of a table is trapezium in shape. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Answer:

The given figure is: 



Lengths of the parallel sides are 1.2 m and 1 m and the perpendicular distance between them is 0.8 m.∴ Area of the trapezium shaped surface=12×(Sum of the parallel sides)×(Perpendicular distance)=12×(1.2+1)×(0.8)=12×2.2×0.8=0.88 m2

Page No 20.23:

Question 11:

The cross-section of a canal is a trapezium in shape. If the canal is 10 m wide at the top 6 m wide at the bottom and the area of cross-section is 72 m2 determine its depth.

Answer:

Let the depth of canal be d.Given:Lengths of the parallel sides of the trapezium shape canal are 10 m and 6 m.And, the area of the cross section of the canal is 72 m2.Area of trapezium=12×(Sum of the parallel sides)×(Perpendicular distance between the parallel sides)72=12×(10+6)×(d)72=8×dd = 728 = 9 m

Page No 20.23:

Question 12:

The area of a trapezium is 91 cm2 and its height is 7 cm. If one of the parallel sides is longer than the other by 8 cm, find the two parallel sides.

Answer:

Given: Area of the trapezium = 91 cm2 Height = 7 cmLet the length of the smaller side be x.Then, the length of longer side will be 8 more than smaller side, i.e. 8+x.Area of trapezium=12×(Sum of the parallel sides)×(Height)91=12×[(8+x)+x]×(7)91=72×[8+x+x]91×2=7×[8+2x]We can rewrite it as follows:7×[8+2x]=182[8+2x]=1827=268+2x=262x=26-8=18x=182=9 cm∴ Length of the shorter side of the trapezium = 9 cm And, length of the longer side = 8+x = 8+9 = 17 cm

Page No 20.23:

Question 13:

The area of a trapezium is 384 cm2. Its parallel sides are in the ratio 3 : 5 and the perpendicular distance between them is 12 cm. Find the length of each one of the parallel sides.

Answer:

Given:Area of the trapezium = 384 cm2The parallel sides are in the ratio 3:5 and the perpendicular height between them is 12 cm.Suppose that the sides are in x multiples of each other.Then, length of the shorter side = 3x Length of the longer side = 5xArea of a trapezium=12×(Sum of parallel sides)×(Height)384=12×(3x+5x)×(12)384=122×(8x)384=6×(8x)8x=3846=64x=648=8 cm∴ Length of the shorter side=3×x=3×8=24 cmAnd, length of the longer side=5×x=5×8=40 cm

Page No 20.23:

Question 14:

Mohan wants to buy a trapezium shaped field. Its side along the river is parallel and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Answer:



Given:Area of the trapezium shaped field = 10500 m2It is also given that the length of the side along the river is double the length of the side along the road.Let us suppose the length of the side along the road to be x.Then, the length of the side along the river = 2×x=2xAnd, the perpendicular distance between these parallel sides =100 mArea of trapezium=12×(Sum of the parallel sides)×(Perpendicular distance)10500=12×(2x+x)×(100)10500=50×(3x)3x=1050050=210x=2103=70 m∴ Length of the side along the river = 2×x = 2×70=140 m



Page No 20.24:

Question 15:

The area of a trapezium is 1586 cm2 and the distance between the parallel sides is 26 cm. If one of the parallel sides is 38 cm, find the other.

Answer:

Given:Area of the trapezium = 1586 cm2Distance between the parallel sides = 26 cmAnd, length of one parallel side = 38 cmLet us suppose the length of the other side to be x cm.Now, area of the trapezium=12×(Sum of the parallel sides)×(Distance between the parallel sides)1586 12×(38+x)×(26)1586=262×(38+x)13×(38+x)=158638+x=158613=122x=122-38=84 cmHence, the length of the other parallel side is 84 cm.

Page No 20.24:

Question 16:

The parallel sides of a trapezium are 25 cm and 13 cm; its nonparallel sides are equal, each being 10 cm, find the area of the trapezium.

Answer:

Given:The parallel sides of a trapezium are 25 cm and 13 cm. Its nonparallel sides are equal in length and each is equal to 10 cm.A rough sketch for the given trapezium is given below:


In above figure, we observe that both the right angle trangles AMD and BNC are congruent triangles.AD = BC = 10 cmD= CN = x cmDMA  =CNB  = 90°Hence, the third side of both the triangles will also be equal. AM=BNAlso, MN=13Since AB = AM+MN+NB: 25=AM+13+BNAM+BN=25-13=12 cmOr, BN+BN=12 cm                           (Because AM=BN)2 BN=12BN=122=6 cm AM = BN = 6 cm.Now, to find the value of x, we will use the Pythagoras theorem in the right angle triangle AMD, whose sides are 10, 6 and x.(Hypotenuse)2=(Base)2+(Altitude)2(10)2=(6)2+(x)2100=36+x2x2=100-36=64x=64=8 cmDistance  between the parallel sides = 8 cm Area of trapezium = 12×(Sum of parallel sides)×(Distance between parallel sides)=12×(25+13)×(8)=152 cm2

Page No 20.24:

Question 17:

Find the area of a trapezium whose parallel sides are 25 cm, 13 cm and the other sides are 15 cm each.

Answer:


Given:Parallel sides of a trapezium are 25 cm and 13 cm. Its nonparallel sides are equal in length and each is equal to 15 cm.A rough skech of the trapezium is given below: 


In above figure, we observe that both the right angle triangles AMD and BNC are similar triangles.This is because both have two common sides as 15 cm and the altitude as x and a right angle.Hence, the remaining side of both the triangles will be equal.AM=BNAlso MN=13Now, since AB=AM+MN+NB: 25=AM+13+BNAM+BN=25-13=12 cmOr, BN+BN=12 cm                                        (Because AM=BN)2 BN=12BN=122=6cm AM=BN=6 cmNow, to find the value of x, we will use the Pythagorian theorem in the right angle triangle AMD whose sides are 15, 6 and x. (Hypotenus)2=(Base)2+(Altitude)2(15)2=(6)2+(x)2225=36+x2 x2= 225-36 =189 x = 189 9×21 = 321 cm Distance between the parallel sides=321 cmArea of trapezium=12×(Sum of parallel sides)×(Distance between the parallel sides)=12×(25+13)×(321)=5721cm2

Page No 20.24:

Question 18:

If the area of a trapezium is 28 cm2 and one of its parallel sides is 6 cm, find the other parallel side if its altitude is 4 cm.

Answer:

Given:Area of the trapezium = 28 cm2Length of one of its parallel sides = 6 cmAltitude = 4 cmLet the other side be x cm.Area of trapezium=12×(Sum of the parallel sides)×(Altitude)28=12×(6+x)×(4)28=2×(6+x)6+x=282=14x=14-6=8 cmHence, the length of the other parallel side of the trapezium is 8 cm.

Page No 20.24:

Question 19:

In Fig. 20.38, a parallelogram is drawn in a trapezium, the area of the parallelogram is 80 cm2, find the area of the trapezium.

Answer:

The given figure is:  


From above figure, it is clear that the length of the parallel sides of the trapezium are 22 cm and 10 cm.Also, it is given that the area of the parallelogram is 80 cm2 and its base is 10 cm.We know:Area of parallelogram=Base×Height80 = 10×Height Height = 8010 = 8 cmSo, now we have the distance between the parallel sides of trapezium, which is equal to 8 cm. Area of trapezium=12×(Sum of the parallel sides)×(Distance between the parallel sides)=12×(22+10)×(8)=128 cm2

Page No 20.24:

Question 20:

Find the area of the field shown in Fig. 20.39 by dividing it into a square, a rectangle and a trapezium.

Answer:

The given figure can be divided into a square, a parallelogram and a trapezium as shown in following figure: 



From the above figure:Area of the figure=(Area of square AGFM with sides 4 cm)+(Area of rectangle MEDN with length 8 cm and width 4 cm)+(Area of trapezium NDCB with parallel sides 8 cm and 3 cm and perpendicular height 4 cm)= (4×4)+(8×4)+[12×(8+3)×(4)]= 16+32+22= 70 cm2



Page No 20.28:

Question 1:

Find the area of the pentagon shown in fig. 20.48, if AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm, BF = 5 cm, CG = 7 cm and EH = 3 cm.

Answer:

The given figure is: 


Given:AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cmBF = 5 cm, CG = 7 cm, EH = 3 cm FG = AG - AF = 8 - 5 = 3 cm And, GD = AD - AG = 10 - 8 = 2 cmFrom given figure:Area of Pantagon = (Area of triangle AFB) + (Area of trapezium FBCG) + (Area of triangle CGD) + (Area of triangle ADE)= (12 × AF × BF) + [12 × (BF + CG) × (FG)] + (12 × GD × CG) + (12 × AD × EH)(12 × 5 × 5) + [12 × (5 + 7) × (3)] + (12 × 2 × 7) + (12 × 10 × 3)= (252) + [362] + (142) + (302)=12.5 + 18 + 7 + 15=52.5 cm2

Page No 20.28:

Question 2:

Find the area enclosed by each of the following figures [Fig. 20.49 (i)-(iii)] as the sum of the areas of a rectangle and a trapezium:

Answer:



(i)The given figure can be divided into a rectangle and a trapezium as shown below:  

From the above firgure:Area of the complete figure = (Area of square ABCF)+(Area of trapezium CDEF)=(AB×BC)+[12×(FC+ED)×(Distance between FC and ED)]=(18×18)+[12×(18+7)×(8)]=324+100=424 cm2

(ii)The given figure can be divided in the following manner: 

From the above figure:AB = AC-BC=28-20=8 cmSo that area of the complete figure = (area of rectangle BCDE)+(area of trapezium ABEF)=(BC×CD)+[12×(BE+AF)×(AB)]=(20×15)+[12×(15+6)×(8)]=300+84=384 cm2


(iii)The given figure can be divided in the following manner:

From the above figure:EF = AB = 6 cmNow, using the Pythagoras theorem in the right angle triangle CDE:52 = 42+CE2CE2 = 25-16=9CE = 9 = 3 cmAnd, GD=GH+HC+CD=4+6+4=14 cm Area of the complete figure=(Area of rectangle ABCH)+(Area of trapezium GDEF)=(AB×BC)+[12×(GD+EF)×(CE)]=(6×4)+[12×(14+6)×(3)]=24+30=54 cm2

Page No 20.28:

Question 3:

There is a pentagonal shaped park as shown in Fig. 20.50. Jyoti and Kavita divided it in two different ways.


Find the area of this park using both ways. Can you suggest some another way of finding its areas?

Answer:


A pentagonal park is given below:  




Jyoti and Kavita divided it in two different ways.  (i) Jyoti divided is into two trapeziums as shown below: 

It is clear that the park is divided in two equal trapeziums whose parallel sides are 30 m and 15 m. And, the distance between the two parallel lines: 152=7.5 m∴ Area of the park=2×(Area of a trapazium)=2×[12×(30+15)×(7.5)]=337.5 m2

(ii)Kavita divided the park into a rectangle and a triangle, as shown in the figure.  

Here, the height of the triangle = 30-15=15 m∴ Area of the park=(Area of square with sides 15 cm)+(Area of triangle with base 15 m and altitude 15 m)=(15×15)+(12×15×15)=225+112.5=337.5 m2



Page No 20.29:

Question 4:

Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm, AO = 60 cm and AD = 90 cm.

Answer:

The given polygon is:  



Given:AL=10 cm, AM=20 cm, AN=50 cmAO=60 cm, AD=90 cmHence, we have the following:MO=AO-AM=60-20=40 cmOD=AD-AO=90-60=30 cmND=AD-AN=90-50=40 cmLN=AN-AL=50-10=40 cmFrom given figure:Area of Polygon=(Area of triangle AMF)+(Area of trapezium MOEF)+(Area of triangle EOD)+(Area of triangle DNC)+(Area of trapezium NLBC)+(Area of triangle ALB)=(12×AM×MF)+[12×(MF+OE)×(OM)]+(12×OD×OE)+(12×DN×NC)+[12×(LB+NC)×(NL)]+(12×AL×LB)=(12×20×20)+[12×(20+60)×(40)]+(12×30×60)+(12×40×40)+[12×(30+40)×(40)]+(12×10×30)=200+1600+900+800+1400+150=5050 cm2

Page No 20.29:

Question 5:

Find the area of the following regular hexagon.

Answer:

The given figure is: 

Join QN.



It is given that the hexagon is regular. So, all its sides must be equal to 13 cm.Also, AN = BQQB+BA+AN = QNAN+13+AN = 232AN = 23-13 = 10AN = 102 = 5 cmHence, AN = BQ = 5 cmNow, in the right angle triangle MAN:MN2=AN2+AM2132=52+AM2AM2=169-25=144AM=144=12cm.∴ OM = RP = 2×AM = 2×12 = 24 cmHence, area of the regular hexagon = (area of triangle MON)+(area of rectangle MOPR)+(area of triangle RPQ)=(12×OM×AN)+(RP×PO)+(12×RP×BQ)=(12×24×5)+(24×13)+(12×24×5)=60+312+60=432 cm2



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