How do you prove that sum of the squares on the sides of a rhombus is equal to the sum of squares on its diagonals?

Answer 1

see explanation.

Some of the properties of a rhombus :
#a)# all sides have equal length :
#=> AB=BC=CD=DA#
#b)# the two diagonals bisect each other at right angle:
#=> OA=OC, => AC=2OA=2OC#
Similarly, #OB=OD, => BD=2OB=2OD#

In #DeltaAOB#, as #angleAOB=90^@#,
by Pythagorean theorem, we know that #AB^2=OA^2+OB^2#
Let the sum of squares of the sides be #S_s#,
#=> S_s=4AB^2#

Let the sum of squares of the diagonals be #S_d#,
#S_d# is given by :
#S_d=AC^2+BD^2=(2OA)^2+(2OB)^2#
#=> S_d=4OA^2+4OB^2=4(OA^2+OB^2)=4AB^2#

#=> S_d=S_s# (proved)

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To prove that the sum of the squares on the sides of a rhombus is equal to the sum of squares on its diagonals, let's denote the side length of the rhombus as ( a ) and the lengths of the diagonals as ( d_1 ) and ( d_2 ).

  1. Sum of squares on the sides: The sum of the squares on the sides is ( 4a^2 ).

  2. Sum of squares on the diagonals: The sum of the squares on the diagonals is ( d_1^2 + d_2^2 ).

To prove the given statement, we can use the properties of a rhombus and the Pythagorean theorem.

Consider half of the rhombus formed by one diagonal and two sides, as shown in the figure below:

    A-------D
   /         \
  /           \
B-------------C

Using the Pythagorean theorem:

For triangle ABD: [ AB^2 + BD^2 = AD^2 ] [ a^2 + \left(\frac{d_1}{2}\right)^2 = \left(\frac{d_2}{2}\right)^2 ] ...(i)

For triangle ACD: [ AC^2 + CD^2 = AD^2 ] [ a^2 + \left(\frac{d_2}{2}\right)^2 = \left(\frac{d_1}{2}\right)^2 ] ...(ii)

Adding equations (i) and (ii):

[ 2a^2 + \frac{d_1^2}{4} + \frac{d_2^2}{4} = \frac{d_1^2 + d_2^2}{2} ]

Multiplying both sides by 4:

[ 8a^2 + d_1^2 + d_2^2 = 2d_1^2 + 2d_2^2 ] [ 8a^2 = d_1^2 + d_2^2 ]

Thus, the sum of the squares on the sides ( 4a^2 ) is equal to the sum of the squares on its diagonals ( d_1^2 + d_2^2 ).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

You can prove that the sum of the squares on the sides of a rhombus is equal to the sum of squares on its diagonals by using the properties of a rhombus and basic geometric principles. One way to prove this is by using the Pythagorean theorem. Since a rhombus is a special type of parallelogram, its diagonals bisect each other at right angles, dividing it into four congruent right triangles. By applying the Pythagorean theorem to each of these triangles and then summing up the results, you can demonstrate the equality between the sum of the squares on the sides and the sum of squares on the diagonals of the rhombus.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7