Q:

# The sum of the diagonals of a rhombus is 5√2.The area is 4 cm²What's the perimeter? The result is 2√34Please, show your work!Thanks :)

Accepted Solution

A:
Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$$
$$= \frac{1}{2} BD (AO + OC)$$
$$\frac{1}{2} BD \times AC$$

$$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$$
$$BD \times AC = 8$$
$$Now, AC + \: BD = 5 \sqrt{2}$$

Squaring both sides, we get
$$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$$
$$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$$
$$AC {}^{2} + BD {}^{2} = 50 - 16$$
$$AC {}^{2} + BD {}^{2} = 34$$

In △AOB, we have
$$OA {}^{2} + OB {}^{2} =AB {}^{2}$$
$$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$$
$$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$$
$$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$$
$$34 = 4 AB {}^{2}$$

Square rooting both sides
$$\sqrt{34} = 2 AB$$
$$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$$.

Hope it helps!