Question

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The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

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Answer 1

Answer:

Using the formulas

P=4a

a=p2+q2

2

Solving forP

P=2p2+q2=2·162+302=68cm

Step-by-step explanation:

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Answer 2

Answer:

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Let PQRS be a rhombus, all sides of rhombus has equal length and its diagonal PR and SQ are intersecting each other at a point O. Diagonals in rhombus bisect each other at 90° .

So, PO = (PR/2)

= 16/2

= 8 cm

And, SO = (SQ/2)

= 30/2

= 15 cm

Then, consider the triangle POS and apply the Pythagoras Theorem,

PS2 = PO2 + SO2

PS2 = 82 + 152

PS2 = 64 + 225

PS2 = 289

PS = √289

PS = 17 cm

Hence, the length of side of rhombus is 17 cm

Now,

Perimeter of Rhombus = 4 × Side of the Rhombus

= 4 × 17

= 68 cm

∴ Perimeter of Rhombus is 68 cm.PS2 = 289

PS = √289

PS = 17 cm

Hence, the length of side of rhombus is 17 cm

Now,

Perimeter of rhombus = 4 × side of the rhombus

= 4 × 17

= 68 cm

∴ Perimeter of rhombus is 68 cm.