Mensuration-i (area Of A Trapezium And A Polygon)
Class 8th Mathematics RD Sharma Solution
- A flooring tile has the shape of a parallelogram whose base is 24 cm and the…
- A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in…
- A playground has the shape of a rectangle, with two semi-circles on its smaller…
- A rectangular piece is 20 m long and 15 m wide. From its four corners,…
- The inside perimeter of a running track (shown in Fig. 20.24) is 400 m. The…
- Find the area of Fig. 20.25, in square cm, correct to one place of decimal.…
- The diamerer of a wheel of a bus is 90 cm which makes 315 revolutions per…
- The area of a rhombus is 240 cm^2 and one of the diagonal is 16 cm. Find…
- The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
- The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars…
- Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If…
- The floor of a building consists of 3000 tiles which are rhombus shaped and…
- A rectangular grassy plot is 112 m long and 78 m broad. It has gravel path 2.5…
- Find the area of a rhombus, each side of which measures 20 cm and one of whose…
- The length of a side of a square field is 4 m. What will be the altitude of…
- Find the area of the field in the form of a rhombus, if the length of each…
- The cost of fencing a square field at 60 paise per metre is Rs. 1200. Find the…
- In exchange of a square plot one of whose sides is 84 m, a man wants to buy a…
- The area of a rhombus is 84 m^2 . If its perimeter is 40 m, then find its…
- A garden is in the form of a rhombus whose side is 30 metres and the…
- A field in the form of a rhombus has each side of length 64 m and altitude 16…
- The area of a rhombus is equal to the area of a triangle whose base and the…
- Find the area, in square metres, of the trapezium whose bases and altitudes are…
- Find the area of trapezium with base 15 cm and height 8 cm, if the side…
- Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm…
- Find the height of a trapezium, the sum of the lengths of whose bases (parallel…
- Find the altitude of a trapezium whose area is 65 cm^2 and whose base are 13 cm…
- Find the sum of the lengths of the bases of a trapezium whose area is 4.2 m^2…
- Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm…
- The area of a trapezium is 960 cm^2 . If the parallel sides are 34 cm and 46…
- Find the area of Fig. 20.35 as the sum of the areas of two trapezium and a…
- Top surface of a table is trapezium in shape. Find its area if its parallel…
- The cross-section of a canal is a trapqzium in shape. If the canal is 10 m…
- The area of a trapezium is 91 cm^2 and its height is 7 cm. If one of the…
- The area of a trapqzium is 384 cm^2 . Its parallel sides are in the ratio 3 :…
- Mohan wants to buy a trapezium shaped field. Its side along the river is…
- The area of a trapezium is 1586 cm^2 and the distance between the parallel…
- The parallel sides of a trapezium are 25 cm and 13 cm; its nonparallel sides…
- Find the area of a trapezium whose parallel sides are 25 cm, 13 cm and the…
- If the area of a trapezium is 28 cm^2 and one of its parallel sides is 6 cm,…
- In Fig. 20.38, a parallelogram is drawn in a trapezium, the area of the…
- Find the area of the field shown in Fig. 20.39 by dividing it into a square, a…
- Find the area of the pentagon shown in fig. 20.48, if AD = 10 cm, AG = 8 cm, AH…
- Find the area enclosed by each of the following figures [fig. 20.49 (i)-(ii)]…
- There is a pentagonal shaped park as shown in Fig. 20.50. Jyoti and Kavita…
- Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm.…
- Find the area of the following regular hexagon.
Exercise 20.1
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 parallelogram = 24cm and height of parallelogram = 10cm
Now, we know that area of parallelogram = base × height
Therefore,
Area of 1 tile = 24 × 10 = 240cm2
And, Area of floor 1080m2
We know that 1m = 100cm
So, 1080m2 = 1080 × 100 × 100 cm2
Now,
Number of tiles = 
Number of tiles = (1080 × 100 × 100)/(24 × 10) = 45000
Therefore number of tiles = 45000
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:
Area of the plot = area of the rectangle + area of semi-circle
Radius of semi-circle = 
Area of the rectagular plot = length × breadth = 60 × 28 = 1680 m2
Area of the semi-circular portion = 
= 
= 308 m2
Therefore, the total area of the plot = 1680 + 308 = 1988 m2
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:
Area of the plot = area of the rectangle + 2 × area of one semi-circle
Radius of semi-circle = 
Area of the plot = length × breadth + 
Area of the plot = 36 × 24.5 + 
Area of the plot = 882 + 471.625 = 1353.625 m2
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:
Area of the plot = area of the rectangle - 4 × area of one quadrant
Radius of semi-circle = 3.5 m
Area of four quadrants = area of one circle
Area of the plot = length × breadth + πr2
Area of the plot = 20 × 15 - 
Area of the plot = 300 – 38.5 = 261.5 m2
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:
Perimeter of the inner track = 2× Length of rectangle + perimeter of two semi-circular ends
Perimeter of the inner track = length + length + 2πr
Hence, the radius of inner circle = 35 m
Radius of outer track = radius of inner track + width of the track
Radius of outer track = 35 + 14 = 49m
Length of outer track = 2× Length of rectangle + perimeter of two outer semi-circular ends
Length of outer track = 2× 90 + 2πr
Length of outer track = 
Length of outer track =
Area of inner track = area of inner rectangle + area of two inner semi-circles
Area of inner track = length× breadth + πr2
Area of inner track = 
Area of inner track = 6300 + 3850
Area of inner track = 10150 m2
Area of outer track = area of outer rectangle + area of two outer semi-circles
Breadth of outer track = 35 + 35 +14 + 14 = 98 m
Area of outer track = length× breadth + πr2
Area of outer track = 
Area of outer track = 8820 + 7546
Area of outer track = 16366 m2
Area of path = area of outer track – area of inner track
Area of path = 16366 – 10150 = 6216 m2
Question 6.
Find the area of Fig. 20.25, in square cm, correct to one place of decimal. (Take π =22/7)
Answer:
Area of the figure = area of square + area of semi circle – area of right angled triangle
Area of the figure = side × side + 
Area of the figure = 10 × 10 + 
Area of the figure = 100 + 78.57 - 24
Area of the figure = 100 + 78.57 – 24 = 154.57 cm2
Area of the figure = 154.57 cm2
Question 7.
The diamerer of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in kilometres per hour. (Take π=22/7)
Answer:
Diameter of wheel = 90 cm
Perimeter of wheel = 2πd
Perimeter of wheel = 
Perimeter of wheel = 282.857 cm
Distance covered in 315 revolutions = 282.857× 315 = 89099.955cm
One km = 100000 cm
Distance covered = 
Speed in km per hour = 0.89 × 60 = 53.4 km per hour
Question 8.
The area of a rhombus is 240 cm2 and one of the diagonal is 16 cm. Find another diagonal.
Answer:
Area of rhombus = 
240 = 
= 
Hence, other diameter is 30 cm
Question 9.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Answer:
Area of rhombus =
Area of rhombus = 
Area of rhombus = 
Hence, area of rhombus is 45 cm2
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:
Area of quadrilateral = 
Area of rhombus = 
Area of rhombus = 
Hence, area of quadrilateral is 252 cm2
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:
Side of rhombus = 6 cm
Altitude of rhombus = 4 cm
Since rhombus is a parallelogram, thereore area of parallelogram = base× altitude
Area of parallelogram = 6 × 4 = 24 cm2
Area of parallelogram = area of rhombus
Area of rhombus =
24 = 
d₂ = 
Hence, length of other diagonal of rhombus is 6 cm
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:
Area of rhombus =
Area of rhombus = 
Area of rhombus = 675 cm2
Area of one tile = 675 cm2
Area of 3000 tiles = 675× 3000 = 2025000 cm2
Area of toles in m2 = 
Cost of tiles = 202.5× 4 = Rs 810.0
Question 13.
A rectangular grassy plot is 112 m long and 78 m broad. It has 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:
Inner area of rectangle = length× breadth
Inner area of rectangle = 112× 78 = 8736 m2
Width of path = 2.5 m
Length of outer rectangle = 112 +2.5 + 2.5 = 117 m
Breadth of outer rectangle = 78 +2.5 + 2.5 = 83 m
Outer area of rectangle = length× breadth
= 117 × 83
= 9711 m2
Area of path = outer area of rectangle – inner area of rectangle
Area of path = 9711 – 8736 = 975 m2
Cost of construction for 1 m2 = 4.50 Rs
Cost of construction for 975 m2 = 975(4.50)
= 4387.5 Rs
Question 14.
Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.
Answer:
Length of side of rhombus = 20 cm
Length of one diagonal = 24 cm
In ΔAOB
Using pythagorous theorem:
AB2 = OA2 + OB2
202 - 122 = OB2
OB2 = 400 – 144
OB2 = 256
OB = 16
Hence, length of other diameter = 16 × 2 = 32 cm
Area of rhombus =
Area of rhombus = 
Area of rhombus = 384 cm2
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:
Length of square = 4 m
Area of square = side2
Area of square = 4 × 4 = 16 m2
Area of square = area of rhombus
Area of rhombus = 16 m2
Area of rhombus =
16 = 
Other diagonal of rhombus = 16 m
In ΔAOB
Using pythagorous theorem:
AB2 = OA2 + OB2
AB2 = 82 + 12
AB2 = 65
AB = 
Rhombus is a parallelogram, area of parallelogram = base× altitude
Area of parallelogram = AB × DE
Area of parallelogram = × DE
DE = 
Altitude of Rhombus = 
cm
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:
Side of rhombus = 14 cm
Altitude of rhombus = 16 cm
Rhombus is a type of parallelogram
Area of parallelogram = base × altitude
Area of parallelogram = 14 × 16 = 224 cm2
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:
Perimeter of square field = 
Perimeter of square field = 
Perimeter of square = 4 × side
Side of square = 
Area of square = side2
Area of square = 500 × 500 = 250000 m2
Cost of reaping = 
Cost of reaping is Rs 
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:
Area of square = side2
Area of square = 84 × 84
Area of square = area of rectangle
84× 84 = 144 × width
Width = 
Therefore width of rectangle = 49 m
Question 19.
The area of a rhombus is 84 m2. If its perimeter is 40 m, then find its altitude.
Answer:
Perimeter of rhombus = 4 × side
Side of rhombus = 
Rhombus is a type of parallelogram, Hence area oa rhombus = area of parallelogram
Therefore area of parallelogram = base× altitude
Base× altitude = area of parallelogram
10 × altitude = 84
Altitude of rhombus = 8.4 m
Question 20.
A garden is in the form of a rhombus whose side is 30 metres and the corresponding altitudes is 16 m. Find the cost of levelling the garden at the rate of Rs. 2 per m2.
Answer:
Side of rhombus = 30 m
Altitude of rhombus = 16 m
Rhombus is a type of parallelogram
Area of parallelogram = base × altitude
Area of parallelogram = 30 × 16 = 480 m2
Cost of levelling the garden = area × rate
Cost of levelling the garden = 480 × 2 = 960
Cost of levelling the garden is Rs 960
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:
Side of rhombus = 30 m
Altitude of rhombus = 16 m
Rhombus is a type of parallelogram
Area of parallelogram = base × altitude
Area of parallelogram = 64 × 16 = 1024 m2
Since area of rhombus = area of square
Area of square = side2
Side of a square = 
Side of square = 
Therefore side of square = 32 m
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:
Length of base of triangle = 24.8 cm
Length of altitude of triangle= 16.5 cm
Area of triangle = 
Area of triangle = 
Area of rhombus = 
204.6 = 
= 
Hence, the length of other diagonal is 18.6 cm
Exercise 20.2
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) Area of trapezium = 
Length of bases of trapezium = 12 dm and 20 dm
10 dm = 1 m
Therefore length of bases in m = 1.2 m and 2 m
Similarly length of altitude in m = 1 m
Area of trapezium = 
Area of trapezium = 
m2
(ii) Area of trapezium =
Length of bases of trapezium = 28 cm and 3 dm
10 dm = 1 m
Therefore length of bases in m = 0.28 m and 0.3 m
Similarly length of altitude in m = 0.25 m
Area of trapezium = 
Area of trapezium = 
m2
(iii) Area of trapezium =
Length of bases of trapezium = 8 m and 60 dm
10 dm = 1 m
Therefore length of bases in m = 8 m and 6 m
Similarly length of altitude in m = 4 m
Area of trapezium = 
Area of trapezium = 
m2
(iv) Area of trapezium =
Length of bases of trapezium = 150 cm and 30 dm
10 dm = 1 m
Therefore length of bases in m = 1.5 m and 3 m
Similarly length of altitude in m = 0.9 m
Area of trapezium = 
Area of trapezium = 
m2
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:
Area of trapezium =
Length of bases of trapezium = 15 cm and 9 cm
Similarly length of altitude in m = 8 cm
Area of trapezium = 
Area of trapezium = 
m2
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:
Area of trapezium =
Length of bases of trapezium = 15 cm and 9 cm
Similarly length of altitude in m = 8 cm
Area of trapezium =
Area of trapezium = m2
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:
Area of trapezium =
Length of bases of trapezium = 60 cm and 60 cm
Area of trapezium = 
Length of altitude = 
Therefore length of altitude = 10 cm
Question 5.
Find the altitude of a trapezium whose area is 65 cm2 and whose base are 13 cm and 26 cm.
Answer:
Area of trapezium =
Length of bases of trapezium = 13 cm and 26 cm
Area of trapezium = 
Length of altitude = 
Therefore length of altitude = 3.33 cm
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:
Area of trapezium =
Area of trapezium = 
Sum of parallel sides = 
Therefore sum of length of parallel sides = 3 m
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:
Area of a trapezium ABCD
= area (∆DFA) + area (rectangle DFEC) + area (∆CEB)
= (1/2 × AF × DF) + (FE × DF) + (1/2 × EB × CE)
= (1/2 × AF × h) + (FE × h) + (1/2 × EB × h)
= 1/2 × h × (AF + 2FE + EB)
= 1/2 × h × (AF + FE + EB + FE)
= 1/2 × h × (AB + FE)
= 1/2 × h × (AB + CD) [Opposite sides of rectangle are equal]
= 1/2 × 6 × (15 + 10)
= 1/2 × 6 × 25 = 75
Area of trapezium = 75 cm2
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:
Area of trapezium = 
Area of trapezium = 
= 
Therefore = 24 cm
Question 9.
Find the area of Fig. 20.35 as the sum of the areas of two trapezium and a rectangle.
Answer:
Area of figure = Area of two trapeziums + area of rectangle
Length of rectangle = 50 cm
Breadth of rectangle = 10 cm
Length of parallel sides of trapezium = 30 cm and 10 cm
Distance between parallel sides of trapezium = 
Area of figure = 
Area of figure = 2 × 
Area of figure = 400
= 900 cm2
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:
Length of parallel sides of trapezium = 1.2m and 1m
Distance between parallel sides of trapezium =
Area of trapezium =
Area of trapezium = 
Area of trapezium = 0.88 m2
Question 11.
The cross-section of a canal is a trapqzium 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:
Length of parallel sides of trapezium = 10m and 6m
Distance between parallel sides of trapezium = x meter
Area of trapezium =
72 = 
Therefore depth of river is 9m
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:
Let the length of one parallel side of trapezium = 
Length of other parallel side of trapezium = 
Area of trapezium = 91 cm2
Area of trapezium =
91 = 
Therefore length of one parallel side of trapezium = 9 cm
Length of other parallel side of trapezium = 9+8 = 17 cm
Question 13.
The area of a trapqzium 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:
Let the length of one parallel side of trapezium = 
Length of other parallel side of trapezium = 
Area of trapezium = 384 cm2
Distance between parallel sides = 12 cm
Area of trapezium =
384 = 
Therefore length of one parallel side of trapezium = 3x = 3× 8 = 24 cm
Length of other parallel side of trapezium = 5x = 5× 8 = 40 cm
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:
Let the length of side of trapezium shaped field along road =
Length of other side of trapezium shaped field along road = 
Area of trapezium = 10500 cm2
Distance between parallel sides = 100 m
Area of trapezium =
10500 = 
Therefore length of side along road = 70 m
Length of side along river = 70× 2 = 140 m
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:
Let the length of other parallel side of trapezium = 
Length of one parallel side of trapezium = 
Area of trapezium = 1586 cm2
Distance between parallel sides = 26 cm
Area of trapezium =
1586 = 
Therefore the length of other parallel side of trapezium = 
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:
In ΔCEF
CE = 10 cm and EF = 6cm
Using pythagorous theorem:
CE2 = CF2 + EF2
CF2 = CE2 - EF2
CF2 = 102 - 62
CF2 = 100-36
CF2 = 64
CF = 8 cm
Area of trapezium = area of parallelogram AECD + area of area of triangle CEF
Area of trapezium = 
Area of trapezium = 
Area of trapezium = 104 + 48 = 152 cm2
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:
In ΔCEF
CE = 10 cm and EF = 6cm
Using pythagorous theorem:
CE2 = CF2 + EF2
CF2 = CE2 - EF2
CF2 = 152 - 62
CF2 = 225-36
CF2 = 189
CF = 17 cm
Area of trapezium = area of parallelogram AECD + area of area of triangle CEF
Area of trapezium = 
Area of trapezium = 
Area of trapezium = 221 + 102 = 323 cm2
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:
Let the length of other parallel side of trapezium =
Length of one parallel side of trapezium = 
Area of trapezium = 28 cm2
Length of altitude of trapezium = 4 cm
Area of trapezium =
28 = 
Therefore the length of other parallel side of trapezium = 
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:
In ΔCEF
CE = 10 cm and EF = 6cm
Using pythagorous theorem:
CE2 = CF2 + EF2
CF2 = CE2 - EF2
CF2 = 102 - 62
CF2 = 100-36
CF2 = 64
CF = 8 cm
Area of parallelogram = 80 cm2
Area of trapezium = area of parallelogram AECD + area of area of triangle CEF
Area of trapezium =
Area of trapezium = 
Area of trapezium = 80 + 48 = 128 cm2
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:
Area of given figure = area of square ABCD + area of rectangle DEFG + area of rectangle GHIJ + area of triangle FHI
Area of given figure = 
Area of given figure = 
Area of given figure = 
= 70 cm2
Exercise 20.3
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:
GH = AG – AH = 8 – 6 = 2 cm
HF = AH – AF = 6 – 5 = 1 cm
GD = AD – AG = 10 – 8 = 2 cm
Area of given figure = area of triangle AFB + area of trapezium BCGF + area of triangle CGD + area of triangle AHE + area of triangle EGD
Area of right angled triangle =
Area of trapezium = 
Area of given pentagon = 
Area of given pentagon = 
Area of given pentagon = 
= 52.5 cm2
Therefore area of given pentagon is 70 cm2
Question 2.
Find the area enclosed by each of the following figures [fig. 20.49 (i)-(ii)] as the sum of the areas of a rectangle and a trapezium.
Answer:
Figure (i)
Area of given figure = Area of trapezium + area of rectangle
Area of given figure = 
Area of given figure = 
Area of given figure = 
Therefore area of given figure is 424 cm2
Figure (ii)
Area of given figure = Area of trapezium + area of rectangle
Area of given figure =
Area of given figure = 
Area of given figure = 
Therefore area of given figure is 384 cm2
Figure (iii)
Using pythagorous theorem in the right angled triangle
52 = 42 + x2
x2 = 25 – 16
x2 = 9
x = 3 cm
Area of given figure = Area of trapezium + area of rectangle
Area of given figure =
Area of given figure = 
Area of given figure = 
Therefore area of given figure is 54 cm2
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 area?
Answer:
Area of given figure = Area of trapezium + area of rectangle
Area of Jyoti’s diagram = 
Area of given figure = 
Area of given figure = 337.5 cm2
Therefore area of given figure is 54 cm2
Area of given pentagon = Area of triangle + area of rectangle
Area of given pentagon = 
Area of given pentagon = 
Area of given pentagon = 
Therefore area of given pentagon is 337.5 cm2
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:
AL = 10 cm; AM = 20 cm; AN = 50 cm; AO = 60 cm; AD = 90 cm given
LM = AM – AL = 20 – 10 = 10 cm
MN = AN – AM = 50 – 20 = 30 cm
OD = AD – AO = 90 – 60 = 30 cm
ON = AO – AN = 60 – 50 = 10 cm
DN = OD + ON = 30 + 10 = 40 cm
OM = MN + ON = 30 + 10 = 40 cm
LN = LM + MN = 10 + 30 = 40 cm
Area of given figure = area of triangle AMF + area of trapezium FMNE + area of triangle END + area of triangle ALB + area of trapezium LBCN + area of triangle DNC
Area of right angled triangle =
Area of trapezium =
Area of given hexagon = 
Area of given hexagon = 
Area of given hexagon = 
= 5050 cm2
Therefore area of given hexagon is 5050 cm2
Question 5.
Find the area of the following regular hexagon.
Answer:
NQ = 23 cm given
ND + QC = 23 -13 = 10
ND = QC = 5 cm
OM = 24 cm
Area of given hexagon = 2 times area of triangle MNO + area of rectangle MOPR
Area of right angled triangle =
Area of rectangle = length × breadth
Area of regular hexagon = 
Area of given hexagon = 
= 432 cm2
Therefore area of given hexagon is 432 cm2