RD Sharma 2019 2020 Solutions for Class 8 Maths Chapter 20 - Mensuration I (Area Of A Trapezium And A Polygon)

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RD Sharma 2019 2020 Solutions for Class 8 Maths 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 Maths 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 2019 2020 Book of class 8 Maths Chapter 20 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2019 2020 Solutions. All RD Sharma 2019 2020 Solutions for class 8 Maths 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:

$Given : Base of a flooring tile that is in the shape of a parallelogram = b = 24 cm Corresponding height = h = 10 cm Now, in a parallelogram : Area (A) = Base (b) \times Height (h) ∴ Area of a tile = 24 cm \times 10 cm = 240 {cm}^{2} Now, observe that the area of the floor is 1080 m^{2} . 1080 m^{2} = 1080 \times 1 m \times 1 m = 1080 \times 100 cm \times 100 cm (Because 1 m = 100 cm) = 1080 \times 100 \times 100 \times cm \times cm = 10800000 {cm}^{2} ∴ Number of required tiles = \frac{10800000}{240} = 45000 Hence, we need 45000 tiles to cover the floor .$

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:

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

$Total area of the plot = Area of rectangle ABCD + Area of semicircle with radius (r = \frac{28}{2} = 14 m) ∴ Area of the rectangular plot with sides 60 m and 28 m = 60 \times 28 = 1680 m^{2} . . . (i) And, area of the semicircle with radius 14 m = \frac{1}{2} π \times (14)^{2} = \frac{1}{2} \times \frac{22}{7} \times 14 \times 14 = 308 m^{2} . . . (ii) ∴ Total area of the plot = 1680 + 308 = 1988 m^{2} . . . (from (i) and (ii))$

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:

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

$Thus, the area of the playground will be the sum of the area of a rectangle and the areas of the two semicircles with equal diameter 24.5 m . Now, area of rectangle with length 36 m and width 24.5 m : Area of rectangle = length \times width = 36 m \times 24.5 m = 882 m^{2} Radius of the semicircle = r = \frac{diameter}{2} = \frac{24.5}{2} = 12.25 m ∴ Area of the semicircle = \frac{1}{2} {πr}^{2} = \frac{1}{2} \times \frac{22}{7} \times (12.25)^{2} = 235.8 m^{2} ∴ Area of the complete playground = area of the rectangular ground + 2 \times area of a semicircle = 882 + 2 \times 235.8 = 1353.6 m^{2}$

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:

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

$∴ Area of the remaining part = Area of the rectangular piece - (4 \times Area of a quadrant of radius 3.5 m) Now, area of the rectangular piece = 20 \times 15 = 300 m^{2} And, area of a quadrant with radius 3.5 m = \frac{1}{4} {πr}^{2} = \frac{1}{4} \times \frac{22}{7} \times (3.5)^{2} = \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 = 9.625 m^{2} ∴ Area of the remaining part = 300 - (4 \times 9.625) = 261.5 m^{2}$

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:

$It is given that the inside perimeter of the running track is 400 m . It means the length of the inner track is 400 m . Let r be the radius of the inner semicircles . Observe : Perimeter of the inner track = Length of two straight portions of 90 m + Length of two semicircles ∴ 400 = (2 \times 90) + (2 \times Perimiter of a semicircle) 400 = 180 + (2 \times \frac{22}{7} \times r) 400 - 180 = (\frac{44}{7} \times r) \frac{44}{7} \times r = 220 r = \frac{220 \times 7}{44} = 35 m ∴ Width of the inner track = 2 r = 2 \times 35 = 70 m Since the track is 14 m wide at all places, so the width of the outer track : 70 + (2 \times 14) = 98 m ∴ Radius of the outer track semicircles = \frac{98}{2} = 49 m Area of the outer track = (Area of the rectangular portion with sides 90 m and 98 m) + (2 \times Area of two semicircles with radius 49 m) = (98 \times 90) + (2 \times \frac{1}{2} \times \frac{22}{7} \times 49^{2}) = (8820) + (7546) = 16366 m^{2} And, area of the inner track = (Area of the rectangular portion with sides 90 m and 70 m) + (2 \times Area of the semicircle with radius 35 m) = (70 \times 90) + (2 \times \frac{1}{2} \times \frac{22}{7} \times 35^{2}) = (6300) + (3850) = 10150 m^{2} ∴ Area of the running track = Area of the outer track - Area of the inner track = 16366 - 10150 = 6216 m^{2} And, length of the outer track = (2 \times length of the straight portion) + (2 \times perimeter of the semicircles with radius 49 m) = (2 \times 90) + (2 \times \frac{22}{7} \times 49) = 180 + 308 = 488 m$

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:

$The given figure is:$

$Construction : Connect A to D . Then, we have : Area of the given figure = (Area of rectangle ABCD + Area of the semicircle) - (Area of ∆ AED) . ∴ Total area of the figure = (Area of rectangle with sides 10 cm and 10 cm) + (Area of semicircle with radius = \frac{10}{2} = 5 cm) - (Area of triangle AED with base 6 cm and height 8 cm) = (10 \times 10) + (\frac{1}{2} \times \frac{22}{7} \times 5^{2}) - (\frac{1}{2} \times 6 \times 8) = 100 + 39.3 - 24 = 115.3 {cm}^{2}$

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:

$It is given that the diameter of the wheel is 90 cm . ∴ Radius of the circular wheel, r = \frac{90}{2} = 45 cm . ∴ Perimeter of the wheel = 2 \times π \times r = 2 \times \frac{22}{7} \times 45 = 282.857 cm It means the wheel travels 282.857 cm in a revolution . Now, it makes 315 revolutions per minute . ∴ Distance travelled by the wheel in one minute = 315 \times 282.857 = 89100 cm ∴ Speed = 89100 cm per minute = \frac{89100 cm}{1 minute} Now, we need to convert it into kilometers per hour . ∴ \frac{89100 cm}{1 minute} = \frac{89100 \times \frac{1}{100000} kilometer}{\frac{1}{60} hour} = \frac{89100}{100000} \times \frac{60}{1} \times \frac{kilometer}{hour} = 53.46 kilometers per hour$

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:

$Given : Area of the rhombus = 240 cm Length of one of its diagonals = 16 cm We know that if the diagonals of a rhombus are d_{1} and d_{2}, then the area of the rhombus is given by : Area = \frac{1}{2} (d_{1} \times d_{2}) Putting the given values : 240 = \frac{1}{2} (16 \times d_{2}) 240 \times 2 = 16 \times d_{2} This can be written as follows : 16 \times d_{2} = 480 d_{2} = \frac{480}{16} d_{2} = 30 cm Thus, the length of the other diagonal of the rhombus is 30 cm .$

Page No 20.14:

Question 9:

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

Answer:

$Given : Lengths of the diagonals of a rhombus are 7.5 cm and 12 cm . Now, we know : Area = \frac{1}{2} (d_{1} \times d_{2}) ∴ Area of rhombus = \frac{1}{2} (7.5 \times 12) = 45 {cm}^{2}$

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:

$Given : Diagonal of a quadrilateral shaped field = 24 m Perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m . Now, we know : Area = \frac{1}{2} \times d \times (h_{1} + h_{2}) ∴ Area of the field = \frac{1}{2} \times 24 \times (8 + 13) = 12 \times 21 = 252 m^{2}$

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:

$Given : Side of the rhombus = 6 cm Altitude = 4 cm One of the diagonals = 8 cm Area of the rhombus = Side \times Altitude = 6 \times 4 = 24 {cm}^{2} . . . . . . . . (i) We know : Area of rhombus = \frac{1}{2} \times d_{1} \times d_{2} Using (i) : 24 = \frac{1}{2} \times d_{1} \times d_{2} 24 = \frac{1}{2} \times 8 \times d_{2} d_{2} = 6 cm$

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:

$Given : The floor consist of 3000 rhombus shaped tiles . The lengths of the diagonals of each tile are 45 cm and 30 cm . ∴ Area of a rhombus shaped tile = \frac{1}{2} \times (45 \times 30) = 675 {cm}^{2} ∴ Area of the complete floor = 3000 \times 675 = 2025000 {cm}^{2} Now, we need to convert this area into m^{2} because the rate of polishing is given as per m^{2} . ∴ 2025000 {cm}^{2} = 2025000 \times cm \times cm = 2025000 \times \frac{1}{100} m \times \frac{1}{100} m = 202.5 m^{2} Now, the cost of polishing 1 m^{2} is Rs 4 . ∴ Total cost of polishing the complete floor = 202.5 \times 4 = 810 Thus, the total cost of polishing the floor is Rs 810 .$

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:

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

$Length of the inner rectangular field = 112 - (2 \times 2.5) = 107 m The width of the inner rectangular field = 78 - (2 \times 2.5) = 73 m ∴ Area of the path = (Area of the rectangle with sides 112 m and 78 m) - (Area of the rectangle with sides 107 m and 73 m) = (112 \times 78) - (107 \times 73) = 8736 - 7811 = 925 m^{2} Now, the cost of constructing the path is Rs 4.50 per square meter . ∴ Cost of constructing the complete path = 925 \times 4.50 = Rs 4162.5 Thus, the total cost of constructing the path is Rs 4162.5$

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:

$Given : Side of the rhombus = 20 cm Length of a diagonal = 24 cm We know : If d_{1} and d_{2} are the lengths of the diagonals of the rhombus, then side of the rhombus = \frac{1}{2} \sqrt{d_{1}^{2} + d_{2}^{2}} So, using the given data to find the length of the other diagonal of the rhombus : 20 = \frac{1}{2} \sqrt{24^{2} + d_{2}^{2}} 40 = \sqrt{24^{2} + d_{2}^{2}} Squ a r i ng both sides to get rid of the square root sign : 40^{2} = 24^{2} + d_{2}^{2} d_{2}^{2} =1600-576=1024 d_{2} = \sqrt{1024} =32 cm ∴ A rea of the rhombus= \frac{1}{2} (24 \times 32) = 384 {cm}^{2}$

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:

$Given: Length of the square field = 4 m ∴ A {rea of the square field = 4×4 = 16 m}^{2} Given : Area of the rhombus = Area of the square field Length of one diagonal of the rhombus = 2 m ∴ S ide of the rhombus= \frac{1}{2} \sqrt{d_{1}^{2} + d_{2}^{2}} And, area of the rhombus= \frac{1}{2} \times (d_{1} \times d_{2}) ∴ A rea: 16 = \frac{1}{2} (2 \times d_{2}) d_{2} =16 m Now, we need to find the length of the side of the rhombus. ∴ Side of the rhombus = \frac{1}{2} \sqrt{2^{2} + 16^{2}} = \frac{1}{2} \sqrt{260} = \frac{1}{2} \sqrt{4 \times 65} = \frac{1}{2} ×2 \sqrt{65} = \sqrt{65} m Also, we know: Area of the rhombus = Side × Altitude ∴ 16= \sqrt{65} ×Altitude Altitude= \frac{16}{\sqrt{65}} m$

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:

$Given: Length of each side of a field in the shape of a rhombus = 14 cm Altitude = 16 cm Now, we know: Area of the rhombus = Side×Altitude ∴ {Area of the field = 14×16 = 224 cm}^{2}$

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:

$Given: C ost of fencing 1 metre of a square field = 60 paise And, the total cost of fencing the entire field = Rs 1200 = 1,20,000 paise ∴ Perimeter of the square field = \frac{120000}{60} = 2000 metres Now, perimeter of a square = 4×side For the given square field : 4×Side = 2000 m Side = \frac{2000}{4} = 500 metres ∴ A {rea of the square field = 500×500=250000 m}^{2} {Again, given: Cost of reaping per 100 m}^{2} = 50 paise ∴ C {ost of reaping per 1 m}^{2} = \frac{50}{100} paise ∴ C {ost of reaping 250000 m}^{2} = \frac{50}{100} \times 250000 = 125000 paise Thus, the total cost of reaping the complete square field is 125000 paise, i.e. Rs 1250.$

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:

$Given: Side of the square plot = 84 m Now, the man wants to exchange it with a rectangular plot of the same area with length 144. A {rea of the square plot = 84×84 = 7056 m}^{2} ∴ Area of the rectangular plot = Length×Width 7056 = 144 \times Width \Rightarrow Width= \frac{7056}{144} =49 m Hence, the width of the rectangular plot is 49 m.$

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:

$Given: {Area of the rhombus = 84 m}^{2} P erimeter = 40 m Now, we know: Perimeter of the rhombus = 4×Side ∴ 40 = 4 \times Side Side= \frac{40}{4} =10 m Again, we know: Area of the rhombus = Side×Altitude \Rightarrow 84 = 10 \times Altitude Altitude = \frac{84}{10} = 8.4 m Hence, the altitude of the rhombus is 8.4 m.$

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:

$Given: Side of the rhombus shaped garden = 30 m Altitude = 16 m Now, area of a rhombus = side×altitude {∴ Area of the given garden=30×16=480 m}^{2} {Also, it is given that the rate of levelling the garden is Rs 2 per 1m}^{2} . ∴ T {otal cost of levelling the complete garden of area 480 m}^{2} =480×2= Rs 960$

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:

$Given: Each side of a rhombus shaped field = 64 m Altitude = 16 m We know: Area of rhombus = Side×Altitude ∴ A {rea of the field = 64×16=1024 m}^{2} Given: Area of the square field = Area of the rhombus {We know: Area of a square=(Side)}^{2} ∴ {1024=(Side)}^{2} \Rightarrow Side= \sqrt{1024} =32 m Thus, the side of the square field is 32 m.$

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:

$Given: Area of the rhombus = Area of the triangle with base 24.8 cm and altitude 16.5 cm Area of the triangle = \frac{1}{2} ×base×altitude = \frac{1}{2} {×24.8×16.5=204.6 cm}^{2} {∴ Area of the rhombus = 204.6 cm}^{2} Also, length of one of the diagonals of the rhombus=22 cm We know : Area of rhombus= \frac{1}{2} (d_{1} \times d_{2}) 204.6 = \frac{1}{2} (22 \times d_{2}) 22 \times d_{2} =409.2 d_{2} = \frac{409.2}{22} =18.6 cm Hence, the length of the other diagonal of the rhombus is 18.6 cm.$

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:

$(i) Given: Bases: 12 dm = \frac{12}{10} m = 1.2 m And, 20 dm= \frac{20}{10} m=2 m Altitude = 10 dm = \frac{10}{10} m = 1 m Area of trapezium = \frac{1}{2} ×(Sum of the bases)×(Altitude) = \frac{1}{2} \times (1.2 + 2) m \times (1) m = 1.6 \times m×m {=1.6 m}^{2} (ii) Given: Bases: 28 cm = \frac{28}{100} m = 0.28 m And, 3 dm = \frac{3}{10} m = 0.3 m Altitude = 25 cm = \frac{25}{100} m = 0.25 m Area of trapezium = \frac{1}{2} ×(Sum of the bases)×(Altitude) = \frac{1}{2} \times (0.28 + 0.3) m \times (0.25) m {= 0.0725 m}^{2} (iii) Given: Bases: 8 m And, 60 dm = \frac{60}{10} m = 6 m Altitude = 40 dm = \frac{40}{10} m = 4 m Area of trapezium= \frac{1}{2} ×(Sum of the bases)×(Altitude) = \frac{1}{2} \times (8 + 6) m \times (4) m = 28 \times m×m {=28 m}^{2} (iv) Given: Bases: 150 cm = \frac{150}{100} m = 1.5 m And, 30 dm = \frac{30}{10} m = 3 m Altitude = 9 dm = \frac{9}{10} m = 0.9 m Area of trapezium= \frac{1}{2} ×(Sum of the bases)×(Altitude) = \frac{1}{2} \times (1.5 + 3) m \times (0.9) m = 2.025 \times m×m {=2.025 m}^{2}$

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:

$Given: Lengths of the parallel sides are 15 cm and 9 cm. Height=8 cm Area of trapezium= \frac{1}{2} ×(Sum of the opposite sides)×(Distance between the parallel sides) = \frac{1}{2} \times (15 + 9) \times (8) {=96 cm}^{2}$

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:

$Given: Lengths of the parallel sides are 16 dm and 22 dm. And, height between the parallel sides is 12 dm. Area of trapezium = \frac{1}{2} ×(Sum of the parallel sides)×(Height) = \frac{1}{2} \times (16 + 22) \times (12) {=228 dm}^{2} = 228 \times dm×dm =228× \frac{1}{10} m× \frac{1}{10} m = 2.28 m^{2}$

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:

$Given: Sum of the parallel sides of a trapezium = 60 cm {Area of the trapezium = 600 cm}^{2} Area of trapezium= \frac{1}{2} ×(Sum of the parallel sides)×(Height) On putting the values : 600 = \frac{1}{2} \times 60 \times (Height) 600 = 30 \times (Height) Height = \frac{600}{30} = 20 cm$

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:

$Given: {Area of the trapezium=65 cm}^{2} The lengths of the opposite parallel sides are 13 cm and 26 cm. Area of trapezium= \frac{1}{2} ×(Sum of parallel bases)×(Altitude) On puttin g the values : 65 = \frac{1}{2} \times (13 + 26) \times (Altitude) 65 \times 2 = 39 \times Altitude Altitude= \frac{130}{39} = \frac{10}{3} cm$

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:

$Given: {Area of the trapezium = 4.2 m}^{2} Height = 280 cm = \frac{280}{100} m = 2.8 m Area of trapezium = \frac{1}{2} \times(Sum of the parallel b ases)×(Height) 4.2 = \frac{1}{2} ×(Sum of the parallel bases)×2.8 4.2 \times 2 = (Sum of the parallel bases)×2.8 Sum of the parallel bases = \frac{8.4}{2.8} = 3 m$

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:

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

$(i) Area of trapezium ABCD =(Area of rectangle EFCD)+(Area of triangle AED+Area of triangle BFC) =(10×6)+[(\frac{1}{2} ×AE×ED)+(\frac{1}{2} ×BF×FC)] =60+[(\frac{1}{2} ×AE×6)+(\frac{1}{2} ×BF×6)] =60+[3 AE+3 BF] =60+3×(AE+BF) Here, AE+EF+FB = 15cm A nd EF = 10 cm ∴ AE+10+BF=15 Or, AE+BF=15-10=5 cm Putting this value in the above formula: A {rea of the trapezium=60+3×(5)=60+15=75 cm}^{2}$

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

$Area of trapezium ABCD=(Area of rectangle ABGH)-[(Area of triangle AHD)+(Area of triangle BGC)] = (15 \times 6) - [(\frac{1}{2} \times DH \times 6) + (\frac{1}{2} \times GC \times 6)] = 90 - [3 ×DH + 3 \times GC] = 90 - 3 [DH+GC] Here, HD+DC+CG=15 cm DC=10 cm HD+10+CG=15 HD+GC=15-10=5 cm Putting this value in the above equation: A {rea of the trapezium=90-3(5)=90-15=75 cm}^{2}$

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:

$Given: {Area of the trapezium = 960 cm}^{2} And the length of the parallel sides are 34 cm and 46 cm. Area of trapezium= \frac{1}{2} ×(Sum of the parallel sides)×(Perpendicular distance between the parallel sides) \Rightarrow 960 = \frac{1}{2} ×(34+46)×(Height) \Rightarrow 960 = 40 ×(Height) \Rightarrow Height = \frac{960}{40} = 24 cm$

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:

$The given figure is:$

$In the given figure, we have a rectangle of length 50 cm and width 10 cm, and two similar trapeziums with parallel sides as 30 cm and 10 cm at both ends. Suppose x is the perpendicular distance between the parallel sides in both the trapeziums. We have: Total length of the given figure= Length of the rectangle+ 2×Perpendicular distance between the parallel sides in both the trapeziums 70 = 50 + 2 \times x 2×x=70-50=20 x= \frac{20}{2} =10 cm Now, area of the complete figure=(area of the rectangle with sides 50 cm and 10 cm)+2×(area of the trapezium with parallel sides 30 cm and 10 cm, and height 10 cm) =(50×10)+2×[\frac{1}{2} ×(30+10)×(10)] = 500 + 2 \times [200] = 900 {cm}^{2}$

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:

$The given figure is:$

$Lengths of the parallel sides are 1.2 m and 1 m and the perpendicular distance between them is 0.8 m. ∴ Area of the trapezium shaped surface= \frac{1}{2} ×(Sum of the parallel sides)×(Perpendicular distance) = \frac{1}{2} \times (1.2 + 1) \times (0.8) = \frac{1}{2} ×2.2×0.8 {=0.88 m}^{2}$

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:

$Let the depth of canal be d . Given: Lengths of the parallel sides of the trapezium shape canal are 10 m and 6 m. {And, the area of the cross section of the canal is 72 m}^{2} . Area of trapezium= \frac{1}{2} ×(Sum of the parallel sides)×(Perpendicular distance between the parallel sides) 72 = \frac{1}{2} \times (10+6)× (d) 72 = 8 × d d = \frac{72}{8} = 9 m$

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:

$Given: {Area of the trapezium = 91 cm}^{2} Height = 7 cm Let the length of the smaller side be x . Then, the length of longer side will be 8 more than smaller side, i.e. 8+ x . Area of trapezium= \frac{1}{2} ×(Sum of the parallel sides)×(Height) \Rightarrow 91= \frac{1}{2} ×[(8+x)+x]×(7) \Rightarrow 91= \frac{7}{2} ×[8+x+x] \Rightarrow 91×2=7×[8+2x] We can rewrite it as follows: 7×[8+2x] = 182 \Rightarrow [8+2 x] = \frac{182}{7} = 26 \Rightarrow 8+2 x = 26 \Rightarrow 2 x =26-8=18 \Rightarrow x = \frac{18}{2} =9 cm ∴ L ength of the shorter side of the trapezium = 9 cm And, length of the longer side = 8+x = 8+9 = 17 cm$

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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:

$Given: {Area of the trapezium = 384 cm}^{2} The parallel sides are in the ratio 3:5 and the perpendicular height between them is 12 cm. Suppose that the sides are in x multiples of each other. Then, length of the shorter side = 3x Length of the longer side = 5x Area of a trapezium= \frac{1}{2} ×(Sum of parallel sides)×(Height) \Rightarrow 384 = \frac{1}{2} \times (3x+5x)×(12) \Rightarrow 384= \frac{12}{2} ×(8x) \Rightarrow 384= 6×(8x) \Rightarrow 8 x = \frac{384}{6} =64 \Rightarrow x = \frac{64}{8} =8 cm ∴ Length of the shorter side=3×x=3×8=24 cm And, length of the longer side=5×x=5×8=40 cm$

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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:

$Given: {Area of the trapezium shaped field = 10500 m}^{2} It is also given that the length of the side along the river is double the length of the side along the road. Let us s uppose the length of the side along the road to be x . Then, the length of the side along the river = 2× x = 2 x And, the perpendicular distance between these parallel sides =100 m Area of trapezium= \frac{1}{2} ×(Sum of the parallel sides)×(Perpendicular distance) 10500 = \frac{1}{2} ×(2x+x)×(100) 10500=50×(3x) 3x= \frac{10500}{50} =210 x= \frac{210}{3} =70 m ∴ L ength of the side along the river = 2× x = 2×70=140 m$

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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:

$Given: {Area of the trapezium = 1586 cm}^{2} Distance between the parallel sides = 26 cm And, length of one parallel side = 38 cm Let us suppose the length of the other side to be x cm . Now, area of the trapezium= \frac{1}{2} ×(Sum of the parallel sides)×(Distance between the parallel sides) \Rightarrow 1586 = \frac{1}{2} ×(38+x)×(26) \Rightarrow 1586 = \frac{26}{2} ×(38+x) \Rightarrow 13 ×(38+ x)=1586 \Rightarrow 38+ x = \frac{1586}{13} =122 \Rightarrow x = 122-38=84 cm Hence, the length of the other parallel side is 84 cm.$

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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:

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

$In above figure, we observe that both the right angle trangles AMD and BNC are congruent triangles. AD = BC = 10 cm D = CN = x cm ∠ DMA = ∠ CNB = 90 ° Hence, the third side of both the triangles will also be equal. ∴ AM=BN Also, MN=13 S ince AB = AM+MN+NB: ∴ 25=AM+13+BN AM+BN=25-13=12 cm Or, BN+BN=12 cm (Because AM=BN) 2 BN=12 BN= \frac{12}{2} =6 cm ∴ AM = BN = 6 cm. Now, to find the value of x, we will use the Pythagoras theorem in the right angle triangle AMD, whose sides are 10, 6 and x. {(Hypotenuse)}^{2} {=(Base)}^{2} {+(Altitude)}^{2} {(10)}^{2} {=(6)}^{2} {+(x)}^{2} {100=36+x}^{2} x^{2} =100-36=64 x= \sqrt{64} =8 cm ∴ D istance between the parallel sides = 8 cm ∴ Area of trapezium = \frac{1}{2} ×(Sum of parallel sides)×(Distance between parallel sides) = \frac{1}{2} ×(25+13)×(8) {=152 cm}^{2}$

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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:

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

$In above figure, we observe that both the right angle triangles AMD and BNC are similar triangles. This is because both have two common sides as 15 cm and the altitude as x and a right angle. Hence, the remaining side of both the triangles will be equal. ∴ AM=BN Also MN=13 Now, since AB=AM+MN+NB: ∴ 25=AM+13+BN AM+BN=25-13=12 cm Or, BN+BN=12 cm (Because AM=BN) 2 BN=12 BN= \frac{12}{2} = 6 cm ∴ AM=BN=6 cm Now, to find the value of x, we will use the Pythagorian theorem in the right angle triangle AMD whose sides are 15, 6 and x. {(Hypotenus)}^{2} = (B ase)^{2} + (A ltitude)^{2} (15)^{2} = (6)^{2} + (x) 2 225 = 36 + x^{2} x^{2} = 225 - 36 = 189 ∴ x = \sqrt{189} = \sqrt{9 \times 21} = 3 \sqrt{21} cm ∴ Distance between the parallel sides= 3 \sqrt{21} cm ∴ Area of trapezium= \frac{1}{2} \times(Sum of parallel sides)×(Distance between the parallel sides) = \frac{1}{2} \times(25+13)\times(3 \sqrt{21}) = 57 \sqrt{21} {cm}^{2}$

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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:

$Given: {Area of the trapezium = 28 cm}^{2} L ength of one of its parallel sides = 6 cm Altitude = 4 cm Let the other side be x cm. Area of trapezium= \frac{1}{2} ×(Sum of the parallel sides)×(Altitude) \Rightarrow 28= \frac{1}{2} ×(6+x)×(4) \Rightarrow 28 = 2 \times (6 + x) \Rightarrow 6+x= \frac{28}{2} =14 \Rightarrow x=14-6=8 cm Hence, the length of the other parallel side of the trapezium is 8 cm.$

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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:

$The given figure is:$

$From above figure, it is clear that the length of the parallel sides of the trapezium are 22 cm and 10 cm. {Also, it is given that the area of the parallelogram is 80 cm}^{2} and its base is 10 cm. We know: Area of parallelogram=Base×Height ∴ 80 = 10×Height Height = \frac{80}{10} = 8 cm So, now we have the distance between the parallel sides of trapezium, which is equal to 8 cm. ∴ A rea of trapezium= \frac{1}{2} ×(Sum of the parallel sides)×(Distance between the parallel sides) = \frac{1}{2} ×(22+10)×(8) {=128 cm}^{2}$

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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:

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

$From the above figure: Area of the figure=(Area of square AGFM with sides 4 cm)+(Area of rectangle MEDN with length 8 cm and width 4 cm)+(Area of trapezium NDCB with parallel sides 8 cm and 3 cm and perpendicular height 4 cm) = (4×4)+(8×4)+[\frac{1}{2} ×(8+3)×(4)] = 16+32+22 {= 70 cm}^{2}$

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:

$The given figure is:$

$Given: AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm BF = 5 cm, CG = 7 cm, EH = 3 cm ∴ FG = AG - AF = 8 - 5 = 3 cm And, GD = AD - AG = 10 - 8 = 2 cm From given figure: Area of Pantagon = (Area of triangle AFB) + (Area of trapezium FBCG) + (Area of triangle CGD) + (Area of triangle ADE) = (\frac{1}{2} × AF × BF) + [\frac{1}{2} \times (BF + CG) × (FG)] + (\frac{1}{2} \times GD × CG) + (\frac{1}{2} × AD × EH) = (\frac{1}{2} × 5 × 5) + [\frac{1}{2} \times (5 + 7) × (3)] + (\frac{1}{2} \times 2 × 7) + (\frac{1}{2} × 10 × 3) = (\frac{25}{2}) + [\frac{36}{2}] + (\frac{14}{2}) + (\frac{30}{2}) = 12.5 + 18 + 7 + 15 {=52.5 cm}^{2}$

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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:

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

$From the above firgure: Area of the complete figure = (Area of square ABCF)+(Area of trapezium CDEF) =(AB×BC)+[\frac{1}{2} ×(FC+ED)×(Distance between FC and ED)] =(18×18)+[\frac{1}{2} ×(18+7)×(8)] =324+100 {=424 cm}^{2}$

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

$From the above figure: AB = AC-BC=28-20=8 cm So that area of the complete figure = (area of rectangle BCDE)+(area of trapezium ABEF) =(BC×CD)+[\frac{1}{2} ×(BE+AF)×(AB)] =(20×15)+[\frac{1}{2} ×(15+6)×(8)] =300+84 {=384 cm}^{2}$

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

$From the above figure: EF = AB = 6 cm Now, using the Pythagoras theorem in the right angle triangle CDE: 5^{2} {= 4}^{2} {+CE}^{2} {CE}^{2} = 25-16=9 CE = \sqrt{9} = 3 cm And, GD=GH+HC+CD=4+6+4=14 cm ∴ Area of the complete figure=(Area of rectangle ABCH)+(Area of trapezium GDEF) =(AB×BC)+[\frac{1}{2} ×(GD+EF)×(CE)] =(6×4)+[\frac{1}{2} ×(14+6)×(3)] =24+30 {=54 cm}^{2}$

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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:

$A pentagonal park is given below:$

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

$It is clear that the park is divided in two equal trapeziums whose parallel sides are 30 m and 15 m. And, the distance between the two parallel lines: \frac{15}{2} =7.5 m ∴ Area of the park=2×(Area of a trapazium) = 2 \times [\frac{1}{2} \times (30 + 15) \times (7.5)] {=337.5 m}^{2}$

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

$Here, the height of the triangle = 30-15=15 m ∴ Area of the park=(Area of square with sides 15 cm)+(Area of triangle with base 15 m and altitude 15 m) =(15×15)+(\frac{1}{2} ×15×15) = 225 + 112.5 {=337.5 m}^{2}$

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:

$The given polygon is:$

$Given: AL=10 cm, AM=20 cm, AN=50 cm AO=60 cm, AD=90 cm Hence, we have the following: MO=AO-AM=60-20=40 cm OD=AD-AO=90-60=30 cm ND=AD-AN=90-50=40 cm LN=AN-AL=50-10=40 cm From given figure: Area of Polygon=(Area of triangle AMF)+(Area of trapezium MOEF)+(Area of triangle EOD)+(Area of triangle DNC)+ (Area of trapezium NLBC)+(Area of triangle ALB) =(\frac{1}{2} ×AM×MF)+[\frac{1}{2} \times (MF+OE)×(OM)]+(\frac{1}{2} ×OD×OE)+(\frac{1}{2} ×DN×NC) +[\frac{1}{2} \times (LB+NC)×(NL)]+(\frac{1}{2} \times AL×LB) = (\frac{1}{2} ×20×20)+[\frac{1}{2} \times (20+60)×(40)]+(\frac{1}{2} \times 30 ×60)+(\frac{1}{2} ×40×40) +[\frac{1}{2} \times (30+40)×(40)]+(\frac{1}{2} \times 10 \times 30) =200+1600+900+800+1400+150 {=5050 cm}^{2}$

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Question 5:

Find the area of the following regular hexagon.

Answer:

$The given figure is:$

Join QN.

$It is given that the hexagon is regular. So, all its sides must be equal to 13 cm. Also, AN = BQ QB+BA+AN = QN AN+13+AN = 23 2AN = 23-13 = 10 AN = \frac{10}{2} = 5 cm Hence, AN = BQ = 5 cm Now, in the right angle triangle MAN: {MN}^{2} {=AN}^{2} {+AM}^{2} 13^{2} {=5}^{2} {+AM}^{2} {AM}^{2} =169-25=144 AM= \sqrt{144} =12cm. ∴ OM = RP = 2×AM = 2×12 = 24 cm Hence, area of the regular hexagon = (area of triangle MON)+(area of rectangle MOPR)+(area of triangle RPQ) =(\frac{1}{2} ×OM×AN)+(RP×PO)+(\frac{1}{2} ×RP×BQ) =(\frac{1}{2} ×24×5)+(24×13)+(\frac{1}{2} ×24×5) =60+312+60 {=432 cm}^{2}$

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