Question

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 :)

Answer 1

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!