Question

[Đặng Quốc Thắng]

In a rhombus of area 60cm² , the ratio between two diagonals is 2:5. The perimeter of the rhombus is \(\sqrt{m}\)cm.

What is the value of m ?

Answer: m = ........

Answer 1

[Phan Thanh Tịnh]

Let x and y be the length of 2 diagonals of the rhombus , so the rhombus's area equal : \(\frac{xy}{2}=\frac{\frac{2}{5}y.y}{2}=\frac{1}{5}y^2=60\)(cm²)

=> y = \(\sqrt{60:\frac{1}{5}}=\sqrt{300}\)(cm) ; x = \(\frac{2}{5}\sqrt{300}=\sqrt{48}\)(cm²) .2 half-diagonals are perpendicular , so the length of 1 side of the rhombus is found by using Pythagorean Theorem :

\(\sqrt{\left(\frac{x}{2}\right)^2+\left(\frac{y}{2}\right)^2}=\sqrt{\frac{\left(\sqrt{48}\right)^2}{4}+\frac{\left(\sqrt{300}\right)^2}{4}}=\sqrt{\frac{48+300}{4}}\)= \(\frac{\sqrt{348}}{2}=\frac{\sqrt{m}}{4}\)(cm)

=> m = \(\left(\frac{\sqrt{348}}{2}.4\right)^2=\frac{348}{4}.16=1392\)