Let the diagonals be x and y and so:-
If: x+y = 11.55 then y = 11.55-x
Area: 0.5*x*y = 16.335 => x*(11.55-x) = 16.335*2
So: 11.55x-x^2 = 32.67 => 11.55x-x^2 -32.67 = 0
Solving the above quadratic equation: x = 4.95 or x = 6.6
Therefore by substitution its diagonals are: 4.95 cm and 6.6 cm
Check: 4,95+6.6 = 11.55 cm which is the sum of its diagonals
Check: 0.5*4.95*6.6 = 16.335 square cm which is its area
Two equilateral triangles can form a rhombus- it can also be formed by using a higher number of isosceles triangles.
1 Let the sides be 5x and 12x2 Using Pythagoras: (5x)2+(12x)2 = 17.5523 And so: 25x2+1442 = 308.0025 => 1692 = 308.00254 Divide both sides by 169 and then square root both sides5 Therefore: x = 1.35 so sides are 5*1.35 = 6.75 cm and 12*1.35 = 16.2 cm6 Area = 6.75*16.2 = 109.35 square cm7 Check with Pythagoras: 6.752+16.22 = 308.0025 and its square root is 17.55 cm which is the rectangle's diagonal
diagram
a diagram
inverse linear or quadratic
Let the other diagonal be x If: 0.5*12*x = 30 then x = 60/12 => x = 5 The rhombus has four interior right angle triangles with lengths of 6 cm and 2.5 cm Using Pythagoras each equal sides of the rhombus works out as 6.5 cm Perimeter: 4*6.5 = 26 cm
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
weener
Let the other parallel side be x and then use Pythagoras to find diagonal lengths: Area: 0.5*(25+x)*18 = 315 sq cm Other parallel side: x = (315*2)/18 -25 = 10 cm Two right angle triangles can be formed with bases of 17.5 and heights of 18 cm Using Pythagoras: 17.5square+18squared = 630.25 and its sq rt is 25 to nearest integer Therefore the diagonal lengths are: 25 cm to the nearest whole number
Perimeter = 29 cm so each side is 7.25 cm. The triangle formed by the diagonal and two sides has sides of 7.25, 7.25 and 11.8 cm so, using Heron's formula, its area is 24.9 square cm. Therefore, the area of the rhombus is twice that = 49.7 square cm.
Double the area and find 2 numbers that have a sum of 42.5 and a product of 375 which will work out as 30 and 12.5 by using the quadratic equation formula. Therefore the diagonals are of lengths 30 and 12.5 which will intersect each other half way at right angles forming 4 right angle triangles inside the rhombus with sides of 15 cm and 6.25 cm Using Pythagoras' theorem each out side length of the rhombus is 16.25 cm and so 4 times 16.25 = 65 cm which is the perimeter of the rhombus.
y = ax
Linear
Linear
0.016
Two equilateral triangles can form a rhombus- it can also be formed by using a higher number of isosceles triangles.
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