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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

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Q: What is the area of a rhombus with diagonals of 12 and 5?
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What is the area of a rhombus with 5 centimeter diagonals?

The diagonals of a rhombus cannot be the same size.


What is the area of a rhombus with the diagonals of 12 and 5?

area_rhombus = product_of_diagonals / 2 = 12 x 5 / 2 = 30 units2 [replace "units" by your measurement unit, eg cm]


What is the perimeter of a rhombus when one of its diagonals is greater than the other diagonal by 5 cm with an area of 150 square cm showing key aspects of work?

Let the diagonals be x+5 and x:- If: 0.5*(x+5)*x = 150 sq cm Then: x2+5x-300 = 0 Solving the above by means of the quadratic equation formula: x = +15 Therefore: diagonals are 15 cm and 20 cm The rhombus has 4 interior right angle triangles each having an hypotenuse Dimensions of their sides: 7.5 and 10 cm Using Pythagoras' theorem: 7.52+102 = 156.25 Its square root: 12.5 cm Thus: 4*12.5 = 50 cm which is the perimeter of the rhombus Note: area of any quadrilateral whose diagonals are perpendicular is 0.5*product of their diagonals


What is the area of a rhombus 10 ft by 12 ft?

Rhombus comprises 4 triangles which can be arranged to form 2 rectangles 5' x 6', so total area is 60 ft2.


What is the perimeter of a rhombus when one of its diagonals is 7.5cm and has an area of 37.5 square cm?

Let the other diagonal be x:- If area is: 0.5*x*7.5 = 37.5 Then x is: 37.5/(0.5*7.5) = 10 The rhombus will then have 4 right angles with sides of 5 and 3.75 Using Pythagoras: hypotenuse of each triangle is 6.25 cm Therefore perimeter of the rhombus is: 4*6.25 = 25 cm