# One of the diagonals of a rhombus is 10 cm long & area of the rhombus is 100 sq.cm. Find the length of side of the rhombus

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Let one diagonal be x cm ( which is 10cm) and another diagonal be x cm. When the two diagonals bisect each other, there are 4 right-angled triangles.

As you know area of rhombus= 1/2 diagonal 1 * diagonal 2

and diagonal 1 is 10 cm, so:

area of rhombus= 1/2* 10* x

5x= 100

x= 20 cm (diagonal 2)

You also know that both diagonals are split into 2 equal lengths when they cut, so x/2= 10/2= 5cm and y/2=20/2=10cm.

By Pythagoras theorem, which is (x/2)^2 + (y/2)^2= (side of rhombus)^2

side of rhombus^2= 5^2+10^2

rhombus side^2= 25+100

rhombus side^2=125

Rhombus side= sqrt 125 = 5sqrt5 cm

It is known the formula of the rhombus aria:

A=(D1XD2)/2, where D1,D2=DIAGONALS OF THE RHOMBUS

A=100sq.cm

D1=10

100=(10XD2)/2

200=10XD2

D2=200/10

D2=20

It is also known that the rhombus diagonals are perpendicular and they are splitting one each other in 2 equal segments (AO=OC, BO=OD). Based on that, it results that inside of rhombus will appear 4 right-angle triangles:AOB, BOC, COD, DOA. The right angle is in the vertex O.

From right angle triangle AOB, applying Pitagora theorem:

AB^2=BO^2+OA^2

where BO=5cm and OA=10cm

AB^2=25+100

AB=sq.root125

AB=5sq.root5cm