what is the perimeter of a 11cm x 10cm rectangle answer 42
square root of -96, which is imaginary. No such triangle is possible in this universe.
Area : 44cm² Perimeter : 30cm
11cm x 17.5cm = 192.5 square centimeters which is 29.837559675119 square inches.
The perimeter is the total distance around the outside of a 2D shape. You calculate it by adding all the lengths of the sides together.So, for example.If I had a rectangle, the length was 11cm, and the width was 5cm.To work out the perimeter of this rectangle, I would simply add all the sides together. So, 11cm+11cm+5cm+5cm=33cmSimple! Hope I helped!
88cm2, but it's a rectangle, not a square.
Its base is: 44/4 = 11cm
330cm^2 (three hundred and thirtycentimeterssquared)
66 square centimeters is.
14cm and 11cm what is the missing side using pythagorean theorem
94.985 or approximately 95 square centimeter.
55 square centimeters
pi*112 = 380 square cm
11 cm x 11 cm = 121 cm2
11cm x 5cm = 55cm2.
Assuming the measurements given are the lengths of the sides of a hexagon, the perimeter is 2 + 2 + 2 + 3 + 5 + 11 = 25 cm.
To calculate the perimeter of a rectangle, you need to add together the length of each side of the rectangle. For a rectangle that is 11cm x 4cm, you would add these numbers together, then multiply it by 2 (this is because there are two sides that are 11cm and two sides that are 4cm). Thus: (11+4) x 2 = 30cm.
11cm * pi 34.5575192 centimeters
The circumference of a circle, if the radius is 11cm, is about 69.12cm
Dimensions are given out as length by width 14cm by 1cm 13cm by 2cm 12cm by 3cm 11cm by 4cm 10cm by 5cm 9cm by 6cm 8cm by 7cm And the rest is all repeats of the above.
The area cannot be 33 cm because the area is measured in SQUARE units of length. That is, square centimetres. The question, therefore, is based on a nonsensical premise. Assume though, that the area is 33 cm2. Then, if one side is 11 cm, the other must be 33/11 = 3 cm. Then the perimeter is 2*(11+3) = 2*14 = 28 cm.
It could be 377.0 cm or 383.3 cm depending on which two adjacent sides.