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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