## RD Sharma Solutions Class 8 Chapter 20 Exercise 20.2

**Exercise 20.2**

**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.**

**Soln:**

**(i)Â **To find the area of a trapezium:

Given data:

Bases of a trapezium = 12 dm (1.2 m) and 20 dm (2 m)

Altitude of a trapezium = 10 dm = lm

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

A = 1/2 x (1.2+2)(1)

A =1.6

Therefore, the area of a trapezium =1.6 m^{2}

**(ii)Â **Â To find the area of a trapezium:

Given data:

Bases of a trapeziumÂ =28 cm (0.28 m) and 3 dm (0.3 m)

Altitude of a trapezium= 25 cm Â = 0.25 m

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

A = 1/2 x (0.28+0.3)(0.25)

Therefore, the area of a trapeziumÂ = 0.0725 m^{2}

**(iii)Â **Given data:

Bases of a trapezium= 8m andÂ 60 dm(6m)

Altitude of a trapezium = 40 dm Â = 4 m

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

A = 1/2 x (8+6)(4)

Therefore, the area of a trapeziumÂ = 28 m^{2}

**(iv)** Given data:

Bases of a trapezium: 150cm (1.5 m) and, 30 dm (3m)

Altitude of a trapezium = 9 dm Â = 0.9 m

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

A =Â 1/2 x (1.5+3)(0.9)

Therefore, the area of a trapeziumÂ = 2.025 m^{2}

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

**Soln:**

Given data: The bases of the trapezium are 15 cm and 9 cm.

HeightÂ of a trapezium= 8 cm

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

A =Â 1/2 x (15+9)(8)

Therefore, the area of a trapezium = 96 cm^{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.**

**Soln:**

Given data: The bases of the trapezium are 16 dm and 22 dm.

HeightÂ of a trapezium= 12 dm

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

A =1/2 x (16+22)(12)

A = 228 dm^{2}

To convert in metre:

A = 228 \(\times\) dm \(\times\) dm

= 228 \(\times\) \(\frac{1}{10}\)m \(\times\) \(\frac{1}{10}\)m

Therefore, the area of a trapezium= 2.28 m^{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}.**

**Soln:**

Given data: The Sum of the parallel sides of a trapezium is 60 cm

Area = 600 cm^{2}

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

600 = (1/2) x 60

600 = 30 x h

h = 600/30Â = 20 cm.

Therefore, the height of a trapezium = 20 cm.

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

**Soln:**

Given data: Area of a trapezium = 65 cm^{2}

The bases of the trapezium are 13 cm and 26 cm

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

65 = (1/2) x (13+26) x altitude

130 = 39 x altitude

Altitude = 130/39

Therefore, the altitude of a trapezium = 10/3 cm

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

**Soln:**

**Given:**

Given data: Area of a trapezium = 4.2 m^{2}

Height of a trapezium = 280 cm Â = 2.8 m

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

4.2 = (1/2)x (sum of the bases)x 2.8

The sum of the bases length = 8.4 / 2.8

Therefore, the sum of the length of the bases of a trapezium = 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.**

**Soln:**

Given Data: The Length of the bases of a trapezium are 10 cm and 15 cm.

The distance or altitude of a trapezium = 6 cm

Now, extend the smaller side of the trapezium and draw the perpendiculars from the ends of both sides.

**(i)**Â From the assumed figure, we can write it as,

Area of trapezium ABCD = Area of a rectangle + (Area of two triangles)

Therefore, area of a trapezium ABCD = AED + BFC

A = (10x 6) + [((1/2)xAExED) + ((1/2)xBFxFC)]

It becomes

A = 60 + [ 3 BF +3 AE ]

A = 60 + 3(AE + BF) ….(1)

From the figure, we can say that

AE + EF + FB = 15 cm and EF = 10 cm

Therefore, AE + 10 + BF = 15

So, AE + BF = 15 â€“ 10

AE + BF= 5 cm

By sunstituting the value in the area of trapezium formula (1), we get

Area of the trapezium = 60 + 3 Â (5) = 60 + 15

Therefore, the area of trapezium= 75 cm^{2}

**(ii)** In case 2,

Area of trapezium ABCD = (Area of ABGH) – [(Area of AHD) + (Area of BGC )]

= (15xÂ 6) –Â [((1/2) x DH x 6)+((1/2)xCGx6)]

= 90 – (3Â DH + 3 GC)

= 90 – 3 (DH + GC ) ….(2)

From the given figure,

HD + CG + DC = 15 cm

We know that DC = 10 cm

Substitute the value of DC

HD + 10 + CG = 15

HD + GC = 5 cm

substituting the above value in equation (2), we get

Area of the trapezium = 10 – 3(5) = 90 â€“ 15

Therefore, the area of a 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.**

**Soln:**

Given data:Â Area of a trapezium = 960 cm^{2}

Length of the trapezium bases is 34 cm and 46 cm.

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

960 = (1/2)x (34+46)x altitude

960 = (1/2) x 80 x altitude

Altitude = 960/40 = 24

Therefore, the height of the trapezium is 24 cm.

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

**Soln:**

From the given figure, we can say that it has one rectangle and two trapeziums.

The rectangle has a dimension of 50 cm x 10 cm

Both the trapeziums have parallel bases with a measure of 30 cm and 10 cm.

Let us assume that, “a” is the perpendicular distance between the parallel sides in both the trapeziums.

Also, from the given figure, we can say that,

The total length of the figure = Rectangle Length + 2(Â perpendicular distance between the parallel sides in both the trapeziums).

So, 70 = 50 +2(a)

2a = 70-50

2a =20

a=10 cm

Therefore, the area of the complete figure = area of a rectangle + 2 ( area of a trapezium)

= (50 x 10) + 2[(1/2)(30+10)(10)

= 500 + 2[(1/2)(400)]

=500 +2(200)

= 500+400

=900

Therefore, the area of the complete figure = 900^{2}

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

**Soln:**

From the given figure, we can say that the parallel sides of the trapezium are 1.2 m and the perpendicular distance between is 0.8 m.

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

A = (1/2) (1.2+1)(0.8)

A = (1/2)(2.2)(0.8)

A = 2.2 x 0.4

A = 0.88m^{2}

Therefore, the area of a trapezium is 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} determine its depth.**

**Soln:**

Assume that the depth of the canal be “d

Given data:

The bases of the trapezium shape canal are 10 m and 6 m.

Area of a trapezium canal =72 m^{2}.

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

72 = (1/2)(10+6)x d

72 = (1/2)(16)d

72 = 8 x d

d = 72/8 = 9

Therefore, the depth of the canal = 9 m

**12. The area of a trapezium is 91 c m ^{2} 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.**

**soln:**

Given data:

Area of a trapezium = 91 cm^{2}

Height of a trapezium = 7 cm

Assume that length of the smaller side of a trapezium be “a”.

Therefore, the length of the longer side of a trapezium will be 8 more than the smaller side of a trapezium, it means that 8 + a.

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

91 = (1/2)(a+(8+a))Â Â (7)

91 = (1/2)(8+2a)(7)

91 = (7/2)(8+2a)

182/7= 8+2a

8 + 2a= 26

2a= 26 â€“ 8 =18

a = 18/2 = 9cm

Therefore, the length of the shorter side of the trapezium, a= 9 cm

Length of the longer side of the trapezium (8+a)= 8 + 9 = 17 cm

**13. The area of a trapezium is 384 c m ^{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.**

**Soln:**

Given data:

Area of a trapezium = 384 cm^{2}

The perpendicular distance between the trapezium is 12 cm and the bases are in the ratio of 3:5

Assume thatÂ the sides are in “a” multiples of each other.

Therefore, the length of the shorter side = 3a

Length of the longer side = 5a

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

384 = (1/2)(3a+5a)(12)

384 =6(8a)

384 = 48a

Therefore, a = 384/48 = 8cm

So, a = 8 cm

Hence, the length of the shorter side of a trapezium = 3(a) = 3(8) = 24 cm

he length of the longer side of a trapezium = 5(a) = 5(8) = 40 cm

**14. Mohan wants to buy a trapezium-shaped field. Its side along the river is parallel and twice the side of the road. If the area of this field is 10500 m ^{2} and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.**

**Soln:**

Given that,

Area of the trapeziumÂ field = 10500 m^{2}

The perpendicular distance between these parallel sides of a trapezium is 100 m

Also, given that the side along the river is parallel and twice the side of the road

Consider that the length of the side along the road to be a.

So, the length of the side along the river = 2 . aÂ = 2a

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

10500 =(1/2) (a+2a)(100)

10500 = 50 (3a)

3a =10500/50= 210

3a = 210

a= 210/3

a= 70

Hence, the length of the side along the river = 2 a= 270Â = 140 m.

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

**Soln:**

Given that: Area of a trapezium = 1586 cm^{2}

Distance between the parallel bases = 26 cm

And, length of one parallel base= 38 cm

Assume that the length of the other side to be “a” cm.

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

1586 =(1/2)(38+a)(26)

1586 =(38+a) (13)

38 + a = 1586/13Â = 122

Therefore, a = 122-38

a = 84 cm

So, the length of the other parallel side of a trapezium is 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.**

**Soln :**

Given that, the bases of a trapezium are 25 cm and 13 cm.

If the nonparallel sides are equal in length which is equal to 10 cm.

Then, the rough sketch for the trapezium will be,

From the above figure, it is observed that both the right angle triangles AMD and BNC are congruent triangles.

AD = BC = 10 cm

D = CN = “x” cm

âˆ DMA = âˆ CNB = 90 degree.

Therefore, the third side of both the triangles will be equal.

So, AM = BN

hence, MN = 13

We know that AB = AM + MN + NB

Substitute the value of MN

25 = AM + 13 + BN

AM + BN = 25 â€“ 13

= 12 cm

Also, BN + BN = 12 cm (sinceAM=I3N)

2 BN = 12

BN = 12/2=6 cm

So, AM = BN = 6cm.

Use Pythagoras theorem, to find the value of “a” the right angle triangle AMD, whose sides are 11

We know that,

(Hypotenuse)^{2 }= (Base)^{2 }+ (Altitude)^{2}

(10)^{2 }= (6)^{2 }+ (a)^{2}

100 = 36 + a^{2}

a^{2 }= 100 â€“ 36 = 64

a = \(\sqrt{64}\)

a =8 cm

Therefore, Distance between the parallel bases isÂ 8 cm

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

= (1/2)(25+13)8

= 38 x4

Therefore, the area of a trapezium=152Â 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.**

**Soln :**

Given that, the bases of a trapezium are 25 cm and 13 cm.

If the nonparallel sides are equal in length which is equal to 15 cm.

Then, the rough sketch for the trapezium will be,

From the above figure, it is observed that both the right angle triangles AMD and BNC are congruent triangles.

AD = BC = 15 cm

D = CN = “x” cm

âˆ DMA = âˆ CNB = 90 degree.

Therefore, the third side of both the triangles will be equal.

So, AM = BN

hence, MN = 13

We know that AB = AM + MN + NB

Substitute the value of MN

25 = AM + 13 + BN

AM + BN = 25 â€“ 13

= 12 cm

Also, BN + BN = 12 cm (sinceAM=I3N)

2 BN = 12

BN = 12/2=6 cm

So, AM = BN = 6cm.

Use Pythagoras theorem, to find the value of “a” the right angle triangle AMD, whose sides are 11

We know that,

(Hypotenuse)^{2 }= (Base)^{2 }+ (Altitude)^{2}

(15)^{2 }= (6)^{2 }+ (a)^{2}

225 = 36 + a^{2}

a^{2 }= 225 â€“ 36 = 189

x = \(\sqrt{189}\)

= \(\sqrt{9\times 21}\)

= 3 \(\sqrt{ 21}\) cm

Therefore, Distance between the parallel bases is 3 \(\sqrt{ 21}\) cm

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

= (1/2)(25+13)3 \(\sqrt{ 21}\)

Therefore, the area of a trapezium=\(57\sqrt{21}\) cm^{2}

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

**soln :**

Given that, Area of a trapezium = 28 cm^{2}

Length of one of the bases of a trapezium = 6 cm

Altitude,h = 4 cm

Assume that other side is “a” cm.

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

28 = (1/2)(6+a)(4)

28 = 2(6+a)

6+a =14

a = 14-6

a = 8 cm

Therefore, the length of the other base of the trapezium is 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.**

**Soln:**

From the given figure, the length of the bases of a trapezium are 10 cm and 22 cm

Area of a parallelogram =80 cm^{2Â }

The base of a parallelogram =10 cm

We know that,

Area of parallelogram = Base x Height square units

80 = 10 xÂ H

Therefore, Height = 80/10 = 8 cm

Hence, the distance between the parallel sides of a trapezium is equal to 8 cm.

We know that the area of trapezium =1/2 x (sum of the bases) x (Altitude) Square units

Substitute the given values in the formula,

A = (1/2)(22+10)(8)

A = (1/2)(32)8

A = 32 x 4

A = 128

Therefore, the area of a trapezium = 128Â 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.**

**Soln:Â **

From the above-given figure, it is divided into a square, a parallelogram and a trapeziumÂ

Therefore, the area of a given 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

A = (4 xÂ 4) + ( 8 xÂ 4) + [(1/2)(8+3)(4)]

A = 16 + 32 + 22

Therefore, the area of the given figure, A= 70 cm^{2}