Câu hỏi của Đặng Quốc Thắng

Question

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 = ........

Phan Thanh Tịnh · Accepted Answer

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 = $\left(\frac{\sqrt{348}}{2}.4\right)^2=\frac{348}{4}.16=1392$ Câu trả lời: 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 = $\left(\frac{\sqrt{348}}{2}.4\right)^2=\frac{348}{4}.16=1392$

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP