Also, lb = 20.
(l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81
⟹ (l + b) = 9.
∴ Perimeter = 2(l + b) = 18 cm.
Perimeter = Distance covered in 8 min. = | ❨ | 12000 | x 8 | ❩m = 1600 m. |
60 |
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
∴ Length = 480 m and Breadth = 320 m.
∴ Area = (480 x 320) m2 = 153600 m2.
Then, AB + BC = 2x metres.
AC = √2x = (1.41x) m.
Saving on 2x metres = (0.59x) m.
Saving % = | ❨ | 0.59x | x 100 | ❩% | = 30% (approx.) |
2x |
∴ A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2.
∴ Percentage error = | ❨ | 404 | x 100 | ❩% | = 4.04% |
100 x 100 |
So, b = 34 ft.
∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
5 | 1 |
4 |
13 | 1 |
2 |
Other side | = |
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ft | ||||||||||||
= |
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ft | |||||||||||||
= |
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ft | |||||||||||||
= | 6 ft. |
∴ Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.
Area of the lawn = 2109 m2.
∴ Area of the crossroads = (2400 - 2109) m2 = 291 m2.
Let the width of the road be x metres. Then,
60x + 40x - x2 = 291
⟹ x2 - 100x + 291 = 0
⟹ (x - 97)(x - 3) = 0
⟹ x = 3.
Area of each tile = (41 x 41) cm2.
∴ Required number of tiles = | ❨ | 1517 x 902 | ❩ | = 814. |
41 x 41 |
2(l + b) | = | 5 |
b | 1 |
⟹ 2l + 2b = 5b
⟹ 3b = 2l
b = | 2 | l |
3 |
Then, Area = 216 cm2
⟹ l x b = 216
⟹l x | 2 | l | = 216 |
3 |
⟹ l2 = 324
⟹ l = 18 cm.
Solving the two equations, we get: l = 63 and b = 40.
∴ Area = (l x b) = (63 x 40) m2 = 2520 m2.
Area to be plastered | = [2(l + b) x h] + (l x b) |
= {[2(25 + 12) x 6] + (25 x 12)} m2 | |
= (444 + 300) m2 | |
= 744 m2. |
∴ Cost of plastering = Rs. | ❨ | 744 x | 75 | ❩ | = Rs. 558. |
100 |
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