The diagonals of a rhombus intersect at right angles and bisect each other.... this should be the only tip you need. Good luck with your homework.
how bout u be a little bit more blunt and just say the awnser??
A+:70ft^2
60 feet
It is a rhombus with sides of length sqrt(61) feet = approx 7.8 feet.
Area of the rhombus: 0.5*7.5*10 = 37.5 square cm Perimeter using Pythagoras: 4*square root of (3.75^2 plus 5^2) = 25 cm
always
It is not clear what the there numbers are. A rhombus has four sides but all four sides are of equal length so only one number is required for the length of all four sides. It is not possible for the three numbers to be the length of the sides and the two diagonals - the numbers simply do not work out.
Area of the rhombus: 0.5*8*10 = 40 square feet
60 feet
Area of a rhombus= 1/2 (d₁) (d₂); where, d₁ and d₂ are the diagonals. Solution: A=1/2 (10) (12) = 60 feet²
40 sq feet
It is: 0.5*10*14 = 70 square feet
If this is a rhombus then the area is half the product of the diagonals - 10 x 14 = 140. Half of 140 = 70, so the area is 70 square feet.
If both diagonals are 10 units then the rhombus is, in fact, a square. Its area is 50 square units.
40have fun cheating on your homework ha-ha
answer: 10 ft2 formula: a=1/2 x d1 x d2 d1 and d2 represent the two diagonals
It is: 0.5*8*10 = 40 square feet
The area of the rhombus is 40 square feet. To see why, Draw a rectangle encompassing the rhombus with sides parallel to the rhombus' diagonals. The rectangle has dimensions 10 ft X 8 ft = 80 square ft. Using the diagonals as dividers, each quarter of the rectangle is divided into 2 by one of the rhombus' sides. Thus the area of the rhombus is exactly half that of the encompassing rectangle.
Find the area of a rhombs with diagonals that measure 8 and 10.