Question

if the diagonal of a rhoumbos are 18 cm and 24 cm respectively, then it's side is equal to?
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Answer 1

Step-by-step explanation:

d1 =18 cm

d2=24 cm

Area =1/2×d1×d2

=1/2×18×24

= 216cm^2

Area of rhombus=4a (a=side of rhombus)

216cm^2 =4a

a=216/4

a = 56 cm

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Answer 2

Answer:

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\linethickness{0.4mm}\put(8.6,3){\large\sf{A}}\put(9,1.85){\sf{18 cm}}\put(10.5,1.55){\sf{24 cm}}\put(7.7,0.9){\large\sf{B}}\put(11.1,0.9){\large\sf{C}}\put(9.9,2.1){\large\sf{O}}\put(8,1){\line(1,0){3}}\qbezier(11,1)(11.5,2)(12,3)\put(9,3){\line(3,0){3}}\qbezier(9,3)(10,2)(11,1)\qbezier(8,1)(10,2)(12,3)\qbezier(8,1)(8.5,2)(9,3)\put(12.1,3){\large\sf{D}}\end{picture}$

• Diagonal₁ = 18 cm
• Diagonal₂ = 24 cm

• Diagonals of Rhombus bisects each other.

⇢ OC = OA = 24/2 cm = 12 cm

⇢ OB = OD = 18/2 cm = 9 cm

⠀

There will be 90° formed at O in all sides, so in ∆OBC

by using Pythagoras Theorem

⇒ BC² = OC² + OB²

⇒ BC² = (12 cm)² + (9 cm)²

⇒ BC² = 144 cm² + 81 cm²

⇒ BC² = 225 cm²

⇒ BC = √225 cm²

⇒ BC = 15 cm

⠀

• BC = CD = DA = AB = 15 cm

⠀

∴ Sides of rhombus will be 15 cm.