find the perimeter of the rombus whose dioganal are 70 cm and 24 cm
Answers
Step-by-step explanation:
we know that
AB = BC = CD = DA In rhombus
The two diagonals are perpendicular and bisect each other
AO = OC = 1/2 AC
BO = OD = 1/2 BD
given that two diagonals are 70 cm and 24 cm
AO = 70/2 = 35 cm BO = 24/2 = 12 cm
from Pythagorean therom we know
AB^2 = AO^2 + BO^2
AB^2 = 35^2 + 12^2
AB^2 = 1225 + 144
AB^2 = 1369
AB = √1368
AB = 37
since AB = BC = CD = DA
Perimeter of rhombus = 4A
perimeter = 4 × 37
perimeter = 148 cn
Answer:
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Step-by-step explanation:
If the length of the diagonals of a rhombus are 70 cm and 24 cm respectively, what is its perimeter?
ABCD is our rhombus with the diagonals
AC=24 cm and BD=70 cm 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 cm and OB=OD=BD/2=70/2=35 cm.
Now, in a rhombus the diagonals intersect each other at 90 degrees.
Hence, ∠BOC=90^o
In right angled triangle BOC, applying the Pythagoras' theorem, we have:
BC=√OB^2+OC^2
⇒BC=√35^2+12^2
⇒BC=√1369
⇒BC=37 cm
Now, in a rhombus, the length of all sides are equal i.e. AB=BC=CD=DA
Thus perimeter = 4⋅(BC) = 4⋅37 = 148 cm.
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