If you mean an area with those measurements as two of the sides - it's 55.4
33 plus 122 plus 459 is 614. 33 plus 122 is 155, and 155 plus 459 is 614.
155% = 1.55 = 1 55/100 = 1 11/20
First find the length of this very very small house and let the length be x: Perimeter = 2(x+25) = 122 inches 2x+50 = 122 2x = 122-50 2x = 72 x = 36 So the length = 36 inches Area = 36*25 = 900 square inches
155 feet = 31 x 5
To find the dimensions of a parking lot with an area of 918 square meters and a perimeter of 122 meters, we can set the length as ( l ) and the width as ( w ). The equations are ( lw = 918 ) and ( 2(l + w) = 122 ). Solving these equations simultaneously, we find that the dimensions are approximately 39 meters in length and 23.5 meters in width.
33 plus 122 plus 459 is 614. 33 plus 122 is 155, and 155 plus 459 is 614.
155% = 1.55 = 1 55/100 = 1 11/20
Let the length be x:- Perimeter = 2(x+25) = 122 inches 2x+50 = 122 2x = 122-50 2x = 72 x = 36 Check: 36+36+25+25 = 122 inches Area = 36*25 = 900 square inches
First find the length of this very very small house and let the length be x: Perimeter = 2(x+25) = 122 inches 2x+50 = 122 2x = 122-50 2x = 72 x = 36 So the length = 36 inches Area = 36*25 = 900 square inches
155 feet = 31 x 5
Let the breadth = xSo, length = 2x+1Thus perimeter will be,2(2x+1+x)=122= 6x+2=122So, 6x=122-2=120So, x= 120/6=20Therefore length=2x+1=2(20)+1=41
Perimeter = 25+36+25+36 = 122 units of measurement Use Pythagoras' theorem to find the other side of the rectangle
My 1987 325ci topped out at 122 when it was new..i'm pretty sure '08s are electronically limited at 155.
To find the dimensions of a parking lot with an area of 918 square meters and a perimeter of 122 meters, we can set the length as ( l ) and the width as ( w ). The equations are ( lw = 918 ) and ( 2(l + w) = 122 ). Solving these equations simultaneously, we find that the dimensions are approximately 39 meters in length and 23.5 meters in width.
1*122=1222*122=2443*122=3664*122=4885*122=6106*122=7327*122=8548*122=9789*122=109810*122=1220...
The maximum area is attained when the rectangle is, in fact, a square. Since the perimeter = 48 feet, the maximum length for a square = 48/4 = 12 feet. So max area = 122 = 144 square feet.
Using Pythagoras' theorem: 162+122 = 400 and the square root of this is 20 (the diagonal) Therefore: 16+12+20 = a perimeter of 48 inches