54
It is: 0.5*10*12 = 60 square m
i think its 288
Since the diagonals of a rhombus are perpendicular and bisect each other, then we can use the Pythagorean theorem to find the length of the side of the rhombus. So in the right triangle, whose length of the legs are 6 and 8 centimeters, the hypotenuse length (the length of the side of the rhombus) is √(62 + 82) = √(36 + 64) = √100 = 10 cm.
The diagonals of a rhombus intersect at 90 degrees therefore it has 4 right angle triangles with sides of 5 and 6 respectively with the hypotenuse being a side of the rhombus. So using Pythagoras' theorem: 52+62 = 61 and the square root of this is the length of each side of the rhombus which is approximately 7.81 units of measurement
it's a 12 sided shape and it has 54 diagonals :D
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.
If you are wanting the area of a rhombus that have diagonals which lengths are 3 and 12, then the area is 18 square units.
60 square units.
Area of a rhombus= 1/2 (d₁) (d₂); where, d₁ and d₂ are the diagonals. Solution: A=1/2 (10) (12) = 60 feet²
60 feet
The diagonals bisect each other. Since that is true then the area of the rhombus is the sum of the two triangles. Half of one diagonal times the other diagonal.2(6x5)/2 or 6x5=30
The lengths of the diagonals work out as 12 cm and 16 cm
It is: 0.5*10*12 = 60 square m
i think its 288
area_rhombus = product_of_diagonals / 2 = 12 x 5 / 2 = 30 units2 [replace "units" by your measurement unit, eg cm]
The area of a rhombus can be calculated using the formula ( \text{Area} = \frac{1}{2} \times d_1 \times d_2 ), where ( d_1 ) and ( d_2 ) are the lengths of the diagonals. In this case, ( d_1 = DF = 9 ) and ( d_2 = EG = 12 ). Therefore, the area is ( \text{Area} = \frac{1}{2} \times 9 \times 12 = 54 ). Thus, the area of rhombus DEFG is 54 square units.
It is a rhombus with sides of length sqrt(61) feet = approx 7.8 feet.