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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
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= sqrt 125 = 5sqrt5 cm
It is known the formula of the rhombus aria:
A=(D1XD2)/2, where D1,D2=DIAGONALS OF THE RHOMBUS
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:
where BO=5cm and OA=10cm
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