Page 269 - Start Up Mathematics_8 (Non CCE)

Now, Perimeter of the rectangle = Perimeter of the square
fi 2(40 + x) = 120
120
fi 40 + x = = 60 fi 40 + x = 60 Puzzle
2
fi x = 60 – 40 = 20 How can you place 15 coins
\ Breadth of the rectangle (b) = 20 cm inside six rectangles such that
Area of rectangle (A ) = l × b = 40 × 20 = 800 cm 2 there are odd numbers of
coins inside each rectangle?
1
2
2
Area of square (A ) = (Side) = (30) = 900 cm 2
2
\ A > A 1
2
2
A – A = (900 – 800) cm = 100 cm 2
2
1
2
Hence, the area of the square is more than that of the rectangle by 100 cm .
2
Example 2: The length and breadth of a rectangular field are in the ratio 4 : 3. Its area is 6,912 m . Find
the cost of fencing the field at ` 2.50 per metre.
Solution: Let the length of the field (l) be 4x m and the breadth (b) be 3x m.
Area of the field = 6,912 m 2
fi l × b = 6,912
fi 4x × 3x = 6,912 fi 12x = 6,912
2
6 912,
2
fi x = = 576 fi x = 24
12
\ Length (l) = 4x = 4 × 24 = 96 m and breadth (b) = 3x = 3 × 24 = 72 m
Now, perimeter of the field = 2(l + b) = 2 × (96 + 72) = 2 × 168 = 336 m

Cost of fencing 1 m = ` 2.50
\ Cost of fencing 336 m = ` (336 × 2.50) = ` 840
Example 3: A rectangular lawn is 30 m × 20 m. It has two paths each 2 m wide running in the middle of
it, one parallel to the length and the other parallel to the breadth. Find the area of the paths.
Solution: Let PQRS and KLMN be the two paths. K 2 m L
Area of PQRS = PQ × PS = 30 m × 2 m = 60 m 2

Area of KLMN = KN × KL = 20 m × 2 m = 40 m 2
P Q
The shaded portion is the common part of the 2 m R 20 m
two paths. S
Area of the shaded portion = 2 m × 2 m = 4 m 2
N M
Hence, the area of the two paths 30 m
= Area of PQRS + area of KLMN – Area of shaded portion
2
2
2
= 60 m + 40 m – 4 m = 96 m 2
Example 4: A square of side 4 cm is inscribed in a circle as shown T 4 cm
in the given figure. Find the area of the shaded portion. R
(Use p = 3.14 and 2 = 1.41) r
Solution: Length of the side of the square = 4 cm
O
2
2
2
In D RUE, RE = RU + EU (Pythagoras Theorem)
2
2
= (4) + (4) = 16 + 16 = 32 E U
RE = 32 = 42 cm
261