If lenght of a diagonal of a rhombus is 30cm nd its area is 240sqcm,Find its perimeter. About the author Abigail
Answer: 34cm. Step-by-step explanation: Let the length of other diagonal of the rhombus be x. Area of the rhombus = 1/2(product of diagonals) => 240cm² = 1/2(30×x) => 240cm² = 30x/2 => 240cm² = 15x => x = 240cm²/15 => x = 16cm. Now, [tex] perimeter \: of \: rhombus \: \\ = > \sqrt{( {d1})^{2} + ( {d2})^{2} } [/tex] [tex] = > \sqrt{( {16})^{2} + ( {30})^{2} } \\ = > \sqrt{256 + 900} \: \: \: \: \: \: \: \\ = > \sqrt{1156} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ = > 34cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex] Hope it helps you. Reply
Answer:
34cm.
Step-by-step explanation:
Let the length of other diagonal of the rhombus be x.
Area of the rhombus = 1/2(product of diagonals)
=> 240cm² = 1/2(30×x)
=> 240cm² = 30x/2
=> 240cm² = 15x
=> x = 240cm²/15
=> x = 16cm.
Now,
[tex] perimeter \: of \: rhombus \: \\ = > \sqrt{( {d1})^{2} + ( {d2})^{2} } [/tex]
[tex] = > \sqrt{( {16})^{2} + ( {30})^{2} } \\ = > \sqrt{256 + 900} \: \: \: \: \: \: \: \\ = > \sqrt{1156} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ = > 34cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
Hope it helps you.