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Assuming that the 40 and 30 refer to sides of a rectangle and not to any of the infinite number of other possible shapes, the answer is 50.
The perimeter of a rectangle measuring 20x45 centimeters is (2*20) + (2*45) = 40 + 90 = 130 cm
Let the width be x: 2(length+width) = perimeter 2(20+x) = 50 40+2x = 50 2x = 50 -40 2x = 10 x = 5 centimeters Check: 5+5+20+20 = 50 centimeters
You are describing a cuboid, not a rectangle. A rectangle has only two diminutions, length and width.
Sqrt(282 + 402) = a whisker under 48 ft 10 in
Area rectangle: 20 times 40 = 800 square inches
Using Pythagoras' theorem it works out as 20 times the square root of 5 which is about 44.721 feet to three decimal places.
46.648 ft
Assuming that the 40 and 30 refer to sides of a rectangle and not to any of the infinite number of other possible shapes, the answer is 50.
The perimeter of a rectangle measuring 20x45 centimeters is (2*20) + (2*45) = 40 + 90 = 130 cm
Let the width be x: 2(length+width) = perimeter 2(20+x) = 50 40+2x = 50 2x = 50 -40 2x = 10 x = 5 centimeters Check: 5+5+20+20 = 50 centimeters
2 x 10 = 20 2 x 20 = 40 add them up 60km
If the length of a rectangle is 12 and the width of the rectangle is 16, by the Pythagorean theorem we know that one diagonal is 20 units long. You can draw two diagonals within a rectangle, so the length of both diagonals together is 20+20 = 40 units.
You are describing a cuboid, not a rectangle. A rectangle has only two diminutions, length and width.
Sqrt(282 + 402) = a whisker under 48 ft 10 in
Let x be the unknown dimension x+x+x+20+x+20 = perimeter of the rectangle 4x+40 = 160 4x = 160-40 4x = 120 x = 30 Therefore the width is 30 cm and the length is 50 cm
800 sq ft