# Class 8 RD Sharma Solutions – Chapter 20 Area Of Trapezium And Polygon- Exercise 20.2 | Set 2

### Chapter 20 Area Of Trapezium And Polygon- Exercise 20.2 | Set 1

### 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 m^{2} determine its depth.

**Solution:**

Given:Length of the parallel sides of the trapezium = 10m and 6m,

Area = 72 m

^{2}Assume that the distance between parallel sides of trapezium is x m

As we know that,

Area of trapezium = 1/2 (Sum of lengths of parallel sides) Ã— distance between parallel sidesNow, put all the given values in this formula, and we get,

72 = 1/2 (10 + 6) Ã— x

72 = 8 Ã— x

x = 72/8 = 9

Hence, the depth is 9m.

### Question 12. The area of a trapezium is 91 cm^{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.

**Solution: **

Given:Assume that the length of one parallel side of trapezium = x m,

then, the length of other parallel side of trapezium = (x+8) m,

Area of trapezium = 91 cm

^{2},Height = 7 cm.

As we know that,

Area of trapezium = 1/2 (Sum of lengths of parallel sides) Ã— altitudeNow, put all the given values in this formula, and we get,

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 – 4 = 9

Hence, the length of one parallel side of trapezium = 9 cm

Therefore the length of other parallel side of trapezium = x+8 = 9+8 = 17 cm.

### Question 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:Assume that the length of one parallel side of trapezium = 3x m,

then the length of other parallel side of trapezium will 5x m,

Area of trapezium = 384 cm

^{2},Distance between the parallel sides of the trapezium = 12 cm.

As we know that,

Area of trapezium = 1/2 (Sum of lengths of parallel sides) Ã— distance between parallel sidesNow, put all the given values in this formula, and we get,

384 = 1/2 (3x + 5x) Ã— 12

384 = 1/2 (8x) Ã— 12

4x = 384/12

4x = 32

x = 8

Hence, the 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.

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

**Solution:**

Given:Assume that the length of side of trapezium shaped field along road = x m

and length of the other side of trapezium shaped field along road = 2x m,

Area of trapezium = 10500 cm

^{2},Distance between the parallel sides of the trapezium = 100 m.

As we know that,

Area of trapezium = 1/2 (Sum of lengths of parallel sides) Ã— distance between parallel sidesNow, put all the given values in this formula, and we get,

10500 = 1/2 (x + 2x) Ã— 100

10500 = 1/2 (3x) Ã— 100

3x = 10500/50

3x = 210

x = 210/3 = 70

x = 70

Hence, the length of side of trapezium shaped field along road is 70 m

and length of other side of trapezium shaped field along road will 2x = 70Ã— 2 = 140 m.

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

**Solution: **

Given:Let us assume that the length of other parallel side of trapezium = x cm

and Length of one parallel side of trapezium = 38 cm,

Area of trapezium = 1586 cm

^{2},Distance between parallel sides = 26 cm.

As we know that,

Area of trapezium = 1/2 (Sum of lengths of parallel sides) Ã— distance between parallel sidesNow, put all the given values in this formula, and we get,

1586 = 1/2 (x + 38) Ã— 26

1586 = (x + 38) Ã— 13

(x + 38) = 1586/13

x = 122 – 38

x = 84

Hence, the length of the other parallel side of the trapezium is 84 cm.

### 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.

**Solution:**

Given:In Î”CEF,

CE = 10 cm and EF = 6cm

By using Pythagoras theorem,

CE

^{2}= CF^{2}+ EF^{2}CF

^{2}= CE^{2}– EF^{2}CF

^{2}= 10^{2}– 6^{2}CF

^{2}= 100 – 36CF

^{2}= 64CF = 8 cm

From the figure we conclude that,

Area of trapezium = Area of parallelogram AECD + Area of triangle CEF= base Ã— height + 1/2 (base Ã— height)

= 13 Ã— 8 + 1/2 (12 Ã— 8) = 104 + 48 = 152

Hence, the area of trapezium is 152 cm^{2}.

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

**Solution:**

Given:In Î”CEF,

CE = 10 cm and EF = 6cm

By using Pythagoras theorem,

CE

^{2}= CF^{2}+ EF^{2}CF

^{2}= CE^{2}– EF^{2}CF

^{2}= 15^{2}– 6^{2}CF

^{2}= 225 – 36CF

^{2}= 189CF = âˆš189 = âˆš (9Ã—21) = 3âˆš21 cm

From the figure we conclude that,

Area of trapezium = Area of parallelogram AECD + Area of triangle CEF= height + 1/2 (sum of parallel sides)

= 3âˆš21 Ã— 1/2 (25 + 13)

= 3âˆš21 Ã— 19 = 57âˆš21

Hence, the area of trapezium is 57âˆš21 cm^{2}.

### Question 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.

**Solution:**

Given:Let us assume that the length of other parallel side of trapezium = x cm,

Length of one parallel side of trapezium = 6 cm,

Area of trapezium = 28 cm

^{2},Length of altitude of trapezium = 4 cm.

As we know that,

Area of trapezium = 1/2 (Sum of lengths of parallel sides) Ã— distance between parallel sidesNow, put all the given values in this formula, and we get,

28 = 1/2 (6 + x) Ã— 4

28 = (6 + x) Ã— 2

(6 + x) = 28/2

(6 + x) = 14

x = 14 – 6 = 8

Hence, the length of the other parallel side of trapezium is 8 cm.

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

**Solution:**

Given:In Î”CEF,

CE = 10 cm and EF = 6cm

By using Pythagoras theorem,

CE

^{2}= CF^{2}+ EF^{2}CF

^{2}= CE^{2}– EF^{2}CF

^{2}= 10^{2}– 6^{2}CF

^{2}= 100-36CF

^{2}= 64CF = 8 cm

Area of parallelogram = 80 cm^2 – (Given)

From the figure we conclude that,

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

So, area of trapezium = base Ã— height + 1/2 (base Ã— height)

Now, put all the given values in this formula, and we get,

= 10 Ã— 8 + 1/2 (12 Ã— 8)

= 80 + 48 = 128

Hence, the area of trapezium is 128 cm^{2}.

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

**Solution:**

From the figure we conclude that,

Area of the given figure = Area of square ABCD + Area of rectangle DEFG +

Area of rectangle GHIJ + Area of triangle FHI

Also,

Area of the given figure = side Ã— side + length Ã— breadth +

length Ã— breadth + 1/2 Ã— base Ã— altitude

Now, put all the given values in this formula, and we get,

= 4Ã—4 + 8Ã—4 + 3Ã—4 + 1/2Ã—5Ã—5

= 16 + 32 + 12 + 10

= 70

Hence, the area of given figure is 70 cm^{2}.

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