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The sum of the diagonals of a rhombus is 5√2.

The area is 4 cm²
What's the perimeter? The result is 2√34

Please, show your work!

Thanks :)

User Focus
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Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD

= (1)/(2) * BD * AO + (1)/(2) * BD * OC

= (1)/(2) BD (AO + OC)

(1)/(2) BD * AC


So, ( BD * AC)/(2) = 4 \: cm {}^(2)

BD * AC = 8

Now, AC + \: BD = 5 √(2)

Squaring both sides, we get

AC {}^(2) + BD {}^(2) + 2 AC.BD =50

AC {}^(2) + BD {}^(2) + 2 * 8 = 50

AC {}^(2) + BD {}^(2) = 50 - 16

AC {}^(2) + BD {}^(2) = 34

In △AOB, we have

OA {}^(2) + OB {}^(2) =AB {}^(2)

( ( AC )/(2) ) {}^(2) + ( ( BD)/(2) ) {}^(2) = AB {}^(2)

\frac{ AC {}^(2) } {4} + \frac{ BD {}^(2) }{4} = AB {}^(2)

AC {}^(2) + BD {}^(2) = 4 AB {}^(2)

34 = 4 AB {}^(2)

Square rooting both sides

√(34) = 2 AB

Perimeter = 4 \: AB \\ = 2 * 2 \: AB \\ = 2 \: * √(34 ) \\ = 2 √(34 \: ) units.

Hope it helps!
The sum of the diagonals of a rhombus is 5√2. The area is 4 cm² What's the perimeter-example-1
User Lucas T
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