Prove that the sum of the the squares of the diagonals of a rhombus is equal to the sum of the squares of its sides
Answers
Step-by-step explanation:
One of the diagonals of a rhombus is equal to one of its sides.
We know that a rhombus has four equal sides. If one side has the same length as a diagonal, the diagonal is part of two equilateral triangles.Since a rhombus has four sides we can say
Rhombus = ABCD where AB = diagonal of AC
Since the triangle ABC is equilateral and so angle ABC = 60. To solve the problem we can make an equation.
60 + x = 180
Solve
60 + x = 180
180 - 60 = x
x = 120
Answer = 120
Answer:
Answer:
Step-by-step explanation:
Given:
A rhombus ABCD
Diagonals AC and BD are perpendicular bisectors
To Prove:
AB² + BC² + CD² + DA² = BD² + AC²
Proof:
In a rhombus we know that the diagonals are perpendicular bisectors.
Hence,
OD = OB = 1/2 BC------(1)
OC = OA = 1/2 AC------(2)
Also in a rhombus, all the 4 sides are equal.
Therefore,
AB = BC = CD = DA------(3)
Now consider ΔODC
By Pythagoras theorem,
DC² = OD² + OC²
Substitute value of OD and OC from equations 1 and 2,
DC² = (1/2 BD)² + (1/2 AC)²
DC² = 1/4 BD² + 1/4 AC²
Multiply the whole equation by 4
4DC² = BD² + AC²
DC² + DC² + DC² + DC² = BD² + AC²
Substitute equation 3 in the above equation,
AB² + BC² + CD² + DA² = BD² + AC²
Hence proved.