Page 175 - Maths Skills - 8

Mensuration 173

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From (i) Area of Trap. ABCD = BE × DF + × CE × DF
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1
= 14 × 6 + × 16 × 6 = 84 + 48 = 132 cm .
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Example 3: Dimensions in the given Fig. are in metres. Find the area of the field.

Solution: Area of field = Area of ∆ABG + Area of ∆BGC + Area of ∆CDF
+ Area of trapezium EDFG + Area of ∆EAG
1 1
Area of ∆ABG = × AG × BG = × 73 × 50 = 1825 m 2 [1]
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1 1
Area of ∆BGC = × (45 + 32) × 50 = × 77 × 50 = 1925 m [2]
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1
Area of ∆CDF = × 32 × 40 = 640 m 2 [3]
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1
1
Area of trapezium EDFG = × (60 + 40) × 45 = × 100 × 45 = 2250 m [4]
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2
2
1
Area of ∆EAG = × 73 × 60 = 2190 m 2 [5]
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Adding (1) to (5) we get
Area of field = (1825 + 1925 + 640 + 2250 + 2190) m = 8830 m .
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Exercise 11.1

1. Find the area of each figure.

(i) (ii) (iii) (iv) (v)

2. One of the parallel side of a trapezium of area 252 cm is 15 cm. The distance between the two parallel
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sides is 14 cm. Find the length of other parallel side.
3. If the height of the parallelogram is 37.3 m and the base is 9.3 m, what is the area of the parallelogram?

4. The area of a trapezium is 177.24 mm . The dimensions of two parallel sides are 28.9 mm and 13.3 mm.
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Find the distance between the parallel sides.
5. What is the area of a square with perimeter 44 cm?
6. The base of the rectangle is 22 cm and the area is 484 cm , what is the height of the rectangle?
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7. If the base of the rectangle is 29.8 mm and the area is 1206.9 mm , what is the perimeter of the rectangle?
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8. Find the area of a parallelogram with base 20 cm, side length 17 cm and height 15 cm.

9. A rectangular garden is 60 m long and 16 m broad. A path of uniform width of 2 m surrounds the
garden inside it. Find the area of the path and the remaining area of the garden. Also, find the cost of
paving the path with bricks at ` 20 per square metre.