What is the perimeter of a rhombus with diagonals 10 and 24?

1 Answer
Nov 15, 2015

The Pythagoras' theorem comes in handy here.

Explanation:

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ABCD is our rhombus with the diagonals AC = 24 units and BD = 10 units respectively.

Now, since a rhombus is also a parallelogram, the diagonals bisect each other at the point where they meet, which in this case is O.

Therefore, OC=OA=(AC)/2=24/2=12 units and OB=OD=(BD)/2=10/2=5 units.

Now, in a rhombus the diagonals intersect each other at 90 degrees.

Hence, angleBOC = 90 ^o
In right angled triangle BOC, applying the Pythagoras' theorem, we have:

BC=sqrt(OB^2+OC^2)
=> BC = sqrt(5^2+12^2)
=>BC = sqrt(169)
=>BC=13 units

Now, in a rhombus, the length of all sides are equal
i.e. AB=BC=CD=DA

Thus perimeter = 4*(BC) = 4*13 = 52 units