This Exercise 20.2 of Chapter 20, deals with the area of a trapezium. Students who find it difficult to solve the problems can refer to RD Sharma Class 8 Solutions which are designed by our expert tutors. Solutions here are prepared in a simple and understandable language which helps students score well in their exams. Students can easily download the pdf from the links provided below and start practising offline.

## Download the Pdf of RD Sharma Solutions for Class 8 Maths Exercise 20.2 Chapter 20 Mensuration – I (Area of a Trapezium and a Polygon)

### Access answers to RD Sharma Maths Solutions For Class 8 Exercise 20.2 Chapter 20 Mensuration – I (Area of a Trapezium and a Polygon)

**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**

**Solution:**

**(i)Â **Given that,

Length of bases of trapezium = 12 dm and 20 dm

Length of altitude = 10 dm

We know that, 10 dm = 1 m

âˆ´ Length of bases in m = 1.2 m and 2 m

Similarly, length of altitude in m = 1 m

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— altitude

Area of trapezium =Â 1/2 (1.2 + 2.0) Ã— 1

Area of trapezium =Â 1/2 Ã— 3.2 = 1.6

So, Area of trapezium =Â 1.6m^{2}

**(ii) **Given that,

Length of bases of trapezium = 28 cm and 3 dm

Length of altitude = 25 cm

We know that, 10 dm = 1 m

âˆ´ Length of bases in m = 0.28 m and 0.3 m

Similarly, length of altitude in m = 0.25 m

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— altitude

Area of trapezium =Â 1/2 (0.28 + 0.3) Ã— 0.25

Area of trapezium =Â 1/2 Ã— 0.58Ã— 0.25 = 0.0725

So, Area of trapezium =Â 0.0725m^{2}

**(iii) **Given that,

Length of bases of trapezium = 8 m and 60 dm

Length of altitude = 40 dm

We know that, 10 dm = 1 m

âˆ´ Length of bases in m = 8 m and 6 m

Similarly, length of altitude in m = 4 m

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— altitude

Area of trapezium =Â 1/2 (8 + 6) Ã— 4

Area of trapezium =Â 1/2 Ã— 56 = 28

So, Area of trapezium =Â 28m^{2}

**(iv) **Given that,

Length of bases of trapezium = 150 cm and 30 dm

Length of altitude = 9 dm

We know that, 10 dm = 1 m

âˆ´ Length of bases in m = 1.5 m and 3 m

Similarly, length of altitude in m = 0.9 m

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— altitude

Area of trapezium =Â 1/2 (1.5 + 3) Ã— 0.9

Area of trapezium =Â 1/2 Ã— 4.5 Ã— 0.9 = 2.025

So, Area of trapezium =Â 2.025m^{2}

**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.**

**Solution:**

Given that,

Length of bases of trapezium = 15 cm and 9 cm

Length of altitude = 8 cm

We know that,

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— altitude

Area of trapezium =Â 1/2 (15 + 9) Ã— 8

Area of trapezium =Â 1/2 Ã— 192 = 96

So, Area of trapezium =Â 96m^{2}

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

**Solution:**

Given that,

Length of bases of trapezium = 16 dm and 22 dm

Length of altitude = 12 dm

We know that, 10 dm = 1 m

âˆ´ Length of bases in m = 1.6 m and 2.2 m

Similarly, length of altitude in m = 1.2 m

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— altitude

Area of trapezium =Â 1/2 (1.6 + 2.2) Ã— 1.2

Area of trapezium =Â 1/2 Ã— 3.8 Ã— 1.2 = 2.28

So, Area of trapezium =Â 2.28m^{2}

**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 cm ^{2}.**

**Solution:**

Given that,

Length of bases of trapezium = 60 cm

Area = 600 cm^{2}

We know that,

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— altitude

600 = 1/2 (60) Ã— altitude

600 = 30 Ã— altitude

Which implies, altitude = 600/30 = 20

âˆ´ Length of altitude is 20 cm

**5.** **Find the altitude of a trapezium whose area is 65 cm ^{2}Â and whose base are 13 cm and 26 cm.**

**Solution:**

Given that,

Length of bases of trapezium = 13 cm and 26 cm

Area = 65 cm^{2}

We know that,

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— altitude

65 = 1/2 (13 + 26) Ã— altitude

65 = 39/2 Ã— altitude

Which implies, altitude = (65Ã—2) /39 = 130/39 = 10/3

âˆ´ Length of altitude = 10/3 cm

**6. Find the sum of the lengths of the bases of a trapezium whose area is 4.2 m ^{2}Â and whose height is 280 cm.**

**Solution:**

Given that,

Height of trapezium = 280 cm = 2.8m

Area = 4.2 m^{2}

We know that,

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— altitudeÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â To calculate the length of parallel sides we can rewrite the above equation as,

Sum of lengths of parallel sides = (2 Ã— Area) / altitude

Sum of lengths of parallel sides = (2 Ã— 4.2) / 2.8 = 8.4/2.8 = 3

âˆ´ Sum of lengths of parallel sides = 3 m

**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.**

**Solution:**

We know that, Area of a trapezium ABCD

= area (âˆ†DFA) + area (rectangle DFEC) + area (âˆ†CEB)

= (1/2 Ã— AF Ã— DF) + (FE Ã— DF) + (1/2 Ã— EB Ã— CE)

= (1/2 Ã— AF Ã— h) + (FE Ã— h) + (1/2 Ã— EB Ã— h)

= 1/2 Ã— h Ã— (AF + 2FE + EB)

= 1/2 Ã— h Ã— (AF + FE + EB + FE)

= 1/2 Ã— h Ã— (AB + FE)

= 1/2 Ã— h Ã— (AB + CD) [Opposite sides of rectangle are equal]

= 1/2 Ã— 6 Ã— (15 + 10)

= 1/2 Ã— 6 Ã— 25 = 75

âˆ´ Area of trapezium = 75 cm^{2}

**8. The area of a trapezium is 960 cm ^{2}. If the parallel sides are 34 cm and 46 cm, find the distance between them.**

**Solution:**

We know that,

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— distance between parallel sides

i.e., Area of trapezium =Â 1/2 (Sum of sides) Ã— distance between parallel sides

To calculate the distance between parallel sides we can rewrite the above equation as,

Distance between parallel sides = (2 Ã— Area) / Sum of sides

= (2 Ã— 960) / (34 + 46)

= (2 Ã— 960) / 80 = 1920/80 = 24

âˆ´ Distance between parallel sides = 24 cm

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

**Solution:**

From the figure we can write,

Area of figure = Area of two trapeziums + Area of rectangle

Given that,

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 = (70â€“50)/2 = 20/2 = 10

So, Distance between parallel sides of trapezium =Â 10 cm

Area of figure = 2 Ã— 1/2 (Sum of lengths of parallel sides) Ã— altitude + Length Ã— Breadth

Area of figure = 2 Ã— 1/2 (30+10) Ã— 10 + 50 Ã— 10

Area of figure = 40 Ã— 10 + 50 Ã— 10

Area of figure = 400 + 500 = 900

âˆ´ Area of figure = 900 cm^{2}

**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.
**

**Solution:**

Given that,

Length of parallel sides of trapezium = 1.2m and 1m

Distance between parallel sides of trapezium = 0.8m

We know that,

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— distance between parallel sides

i.e., Area of trapezium =Â 1/2 (Sum of sides) Ã— distance between parallel sides

Area of trapezium =Â 1/2 (1.2 + 1) Ã— 0.8

Area of trapezium =Â 1/2 Ã— 2.2 Ã— 0.8 = 0.88

So, Area of trapezium =Â 0.88m^{2}

**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 m ^{2}Â determine its depth.**

**Solution:**

Given that,

Length of parallel sides of trapezium = 10m and 6m

Area = 72 m^{2}**Â **

Let the distance between parallel sides of trapezium = x meter

We know that,

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— distance between parallel sides

i.e., Area of trapezium =Â 1/2 (Sum of sides) Ã— distance between parallel sides

72 = 1/2 (10 + 6) Ã— x

72 = 8 Ã— x

x = 72/8 = 9

âˆ´ The depth is 9m.

**12.** **The area of a trapezium is 91 cm ^{2}Â 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.**

**Solution:**

Given that,

Let the length of one parallel side of trapezium =Â x meter

Length of other parallel side of trapezium =Â (x+8) meter

Area of trapezium = 91 cm^{2}

Height = 7 cm

We know that,

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— altitude

91 = 1/2 (x+x+8) Ã— 7

91 = 1/2(2x+8) Ã— 7

91 = (x+4) Ã— 7

(x+4) = 91/7

x+4 = 13

x = 13 â€“ 4

x = 9

âˆ´ Length of one parallel side of trapezium = 9 cm

And, Length of other parallel side of trapezium = x+8 = 9+8 = 17 cm

**13.** **The area of a trapezium is 384 cm ^{2}. 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.**

**Solution:**

Given that,

Let the length of one parallel side of trapezium =Â 3x meter

Length of other parallel side of trapezium =Â 5x meter

Area of trapezium = 384 cm^{2}

Distance between parallel sides = 12 cm

We know that,

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— distance between parallel sides

i.e., Area of trapezium =Â 1/2 (Sum of sides) Ã— distance between parallel sides

384 = 1/2 (3x + 5x) Ã— 12

384 = 1/2 (8x) Ã— 12

4x = 384/12

4x = 32

x = 8

âˆ´ Length of one parallel side of trapezium = 3x = 3Ã— 8 = 24 cm

And, Length of other parallel side of trapezium = 5x = 5Ã— 8 = 40 cm

**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 m ^{2}Â and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
**

**Solution:**

Given that,

Let the length of side of trapezium shaped field along road =Â x meter

Length of other side of trapezium shaped field along road =Â 2x meter

Area of trapezium = 10500 cm^{2}

Distance between parallel sides = 100 m

We know that,

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— distance between parallel sides

i.e., Area of trapezium =Â 1/2 (Sum of sides) Ã— distance between parallel sides

10500 = 1/2 (x + 2x) Ã— 100

10500 = 1/2 (3x) Ã— 100

3x = 10500/50

3x = 210

x = 210/3 = 70

x = 70

âˆ´ Length of side of trapezium shaped field along road = 70 m

And, Length of other side of trapezium shaped field along road = 2x = 70Ã— 2 = 140 m

**15.** **The area of a trapezium is 1586 cm ^{2}Â and the distance between the parallel sides is 26 cm. If one of the parallel sides is 38 cm, find the other.**

**Solution:**

Given that,

Let the length of other parallel side of trapezium =Â x cm

Length of one parallel side of trapezium =Â 38 cm

Area of trapezium = 1586 cm^{2}

Distance between parallel sides = 26 cm

We know that,

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— distance between parallel sides

i.e., Area of trapezium =Â 1/2 (Sum of sides) Ã— distance between parallel sides

1586 = 1/2 (x + 38) Ã— 26

1586 = (x + 38) Ã— 13

(x + 38) = 1586/13

x = 122 â€“ 38

x =84

âˆ´ Length of the other parallel side of trapezium = 84 cm

**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.**

**Solution:**

In Î”CEF,

CE = 10 cm and EF = 6cm

Using Pythagoras theorem:

CE^{2}Â = CF^{2}Â + EF^{2}

CF^{2}Â = CE^{2}Â – EF^{2}

CF^{2}Â = 10^{2}Â – 6^{2}

CF^{2}Â = 100-36

CF^{2}Â = 64

CF = 8 cm

From the figure we can write,

Area of trapezium = Area of parallelogram AECD + Area of area of triangle CEF

Area of trapezium = base Ã— height + 1/2 (base Ã— height)

Area of trapezium =Â 13 Ã— 8 + 1/2 (12 Ã— 8)

Area of trapezium =Â 104 + 48 = 152

âˆ´ Area of trapezium = 152 cm^{2}

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

**Solution:**

In Î”CEF,

CE = 10 cm and EF = 6cm

Using Pythagoras theorem:

CE^{2}Â = CF^{2}Â + EF^{2}

CF^{2}Â = CE^{2}Â – EF^{2}

CF^{2}Â = 15^{2}Â – 6^{2}

CF^{2}Â = 225-36

CF^{2}Â = 189

CF = âˆš189

= âˆš (9Ã—21)

= 3âˆš21 cm

From the figure we can write,

Area of trapezium = Area of parallelogram AECD + Area of area of triangle CEF

Area of trapezium = height + 1/2 (sum of parallel sides)

Area of trapezium =Â 3âˆš21 Ã— 1/2 (25 + 13)

Area of trapezium =Â 3âˆš21 Ã— 19 = 57âˆš21

âˆ´ Area of trapezium = 57âˆš21 cm^{2}

**18.** **If the area of a trapezium is 28 cm ^{2}Â and one of its parallel sides is 6 cm, find the other parallel side if its altitude is 4 cm.**

**Solution:**

Given that,

Let the length of other parallel side of trapezium =Â x cm

Length of one parallel side of trapezium =Â 6 cm

Area of trapezium = 28 cm^{2}

Length of altitude of trapezium = 4 cm

We know that,

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— distance between parallel sides

i.e., Area of trapezium =Â 1/2 (Sum of sides) Ã— distance between parallel sides

28 = 1/2 (6 + x) Ã— 4

28 = (6 + x) Ã— 2

(6 + x) = 28/2

(6 + x) = 14

x = 14 â€“ 6

x = 8

âˆ´ Length of the other parallel side of trapezium = 8 cm

**19. In Fig. 20.38, a parallelogram is drawn in a trapezium, the area of the parallelogram is 80 cm ^{2}, find the area of the trapezium.
**

**Solution:**

In Î”CEF,

CE = 10 cm and EF = 6cm

Using Pythagoras theorem:

CE^{2}Â = CF^{2}Â + EF^{2}

CF^{2}Â = CE^{2}Â – EF^{2}

CF^{2}Â = 10^{2}Â – 6^{2}

CF^{2}Â = 100-36

CF^{2}Â = 64

CF = 8 cm

Area of parallelogram = 80 cm^{2}

From the figure we can write,

Area of trapezium = Area of parallelogram AECD + Area of area of triangle CEF

Area of trapezium = base Ã— height + 1/2 (base Ã— height)

Area of trapezium =Â 10 Ã— 8 + 1/2 (12 Ã— 8)

Area of trapezium =Â 80 + 48 = 128

âˆ´ Area of trapezium = 128 cm^{2}

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

**Solution:**

From the figure we can write,

Area of given figure = Area of square ABCD + Area of rectangle DEFG + Area of rectangle GHIJ + Area of triangle FHI

i.e., Area of given figure = side Ã— side + length Ã— breadth + length Ã— breadth + 1/2 Ã— base Ã— altitude

Area of given figure = 4Ã—4 + 8Ã—4 + 3Ã—4 + 1/2Ã—5Ã—5

Area of given figure =Â 16 + 32 + 12 + 10 = 70

âˆ´ Area of given figure =Â 70 cm^{2}