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Q: What is the perimeter of a rhombus with an area of 42 and an altitude of 6?

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Area 42 cm2, perimeter 26 cm.

Assuming that it's a rectangle then:- Area = 42*14 = 588 square cm Perimeter = 42+42+14+14 = 112 cm

Area is 64 units squared.

The area is about 127.3 square inches.

Perimeter = 168 inches so 4*length of side = 168 inches Length of side = 168/4 = 42 inches So area = 42*42 = 1764 square inches.

That depends how you define the perimeter for a cube. The term perimeter is usually used for plane figures, not for 3D solids.

Even if you knew how many sides the polygon has, you stillcould not calculate its perimeter with that much information.Examples:-- An equilateral triangle with area of 20 has perimeter of 20.3885 .-- A square with area of 20 has perimeter of 17.889(rounded).-- A rectangle with area of 20 can have any perimeter more than 17.889 .4 by 5 . . . . area = 20, perimeter = 182 by 10 . . . area = 20, perimeter = 241 by 20 . . . area = 20, perimeter = 42..etc.

The 2 lengths that you described are diagonals. The area of a rhombus when you know the diagonals is half the product of the diagonals: Area = (1/2) * ( 12 * 7) = 42.

For a given perimeter, the greatest possible area is enclosed by a circle.A circle with a circumference of 18 has a diameter of (18/pi) and a radius of (9/pi).Its area is (pi R2) = (pi 92/pi2) = 81/pi = 25.78 (rounded)So an area of 42 cannot be enclosed by a perimeter of 18.

Largest = 86, Smallest 26

The perimeter of a rectangle is 42. Meters. The length of the rectangle is threemeter less than twice the width.Mar

None. 42 yards is a length with no width. Therefore it cannot have a perimeter.

No! A rhombus has only 4 sides in total!

When the linear dimensions of a plane figure are quadrupled, its perimeter is quadrupled, and its area is multiplied by 42 = 16 .

The dimensions are: altitude 12 inches and base 7 inches Check: 0.5*12*7 = 42 square inches

The perimeter of the rectangle is 42 units

42 square units.

Make it 2 wide and 21 long and you've got it.

Try a rectangle with dimensions of 6 cm and 7 cm, and see what you get.

sum of sides is perimeter:14+14+7+7 = 42 cm

The area doesn't tell you the shape or the perimeter. Even if you restrict it torectangles, there are an infinite number of those, all with different perimeters,that all have 42 ft squared of area.Here are a few of them. ALL of the areas are 42 ft squared.42 ft by 1 ft (perimeter = 86 ft)21 ft by 2 ft (perimeter = 46 ft)14 ft by 3 ft (perimeter = 34 ft)12 ft by 3ft6in (perimeter = 31 ft)7 ft by 6 ft (perimeter = 26 ft)If you're trying to make a square out of it, then the length of each side of the squareis sqrt(42), and the perimeter is 25.92 to the nearest hundredth.On the other hand, it could be a circle, and the perimeter (circumference) of the circleis 22.97 to the nearest hundredth.

12

The dimensions of the rectangle are 3 inches by 14 inches

There is no limit to the size of the perimeter.

Perimeter is the distance AROUND any particular object where area is the amount of space inside the object (2dimentional). Let's say you have a yard 10 feet long, by 11 feet wide. The perimeter is the distance AROUND your yard or 10+10+11+11 = 42ft The AREA is the amount of space inside or L*W or 10*11 = 110ft^2 Suppose I change the dimensions of the yard without changing its perimeter (42 ft) and see what happens to the area. Suppose the length and width are 20ft and 1 ft respectively. The perimeter is still 20+1+20+1=42. The area however is 20X1 =20 ft2 Now if we take the yard to be a square one of 10.5 ft side then the perimeter still remains 42 ft. but the area becomes 110.25 ft2. Now suppose I assume that the perimeter of a circle is 42 ft. What will the area of such a circle be? Now we know that the circumference of a circle is 2x3.14(pi)Xr = 42 ft. Therefore r = 6.687 ft. r2= 44.715 ft2 therefore area of the circle is 4xpixr2= 561.63 ft 2. Thus we can deduce that given a known perimeter a circular shape has or occupies the largest area. What happens if the shape is irregular? Say a irregular polygon? Would my deduction that the cicrular shape has the largest area remain valid?