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Posted by Kashish Taank 8 years, 3 months ago
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Sneha Sneha 8 years, 3 months ago
Sneha Sneha 8 years, 3 months ago
Naveen Sharma 8 years, 3 months ago
Perimeter of Rhombus = 100 m
Side of Rhombus = {tex}{100\over 4}= 25 m {/tex}
One diagonal of Rhombus (d1) = 30 m
Second diagonal of rhombus = d2
As we know, diagonal of rhombus bisect each other at 90 degree. So,
{tex}d_1^2+d_2^2= 4(side)^2\\
=>900 + d_2^2 = 4\times 625\\
=> d_2^2 = 2500-900=1600\\
=> d_2 = 40{/tex}
Area of rhombus = {tex}{1\over 2}\times d_1\times d_2\\ = {1\over 2}\times 30\times 40 \\ = 600 m^2{/tex}
Posted by Laxmi Nair 8 years, 3 months ago
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Golua Singham 8 years, 3 months ago
Varun Banda 8 years, 3 months ago
Given : Let ABCD is cyclic quadrilateral.
To prove : ∠A + ∠C = 180° and ∠B + ∠D = 180°.
Construction: join OB and OD.
Proof :
∠BOD = 2 ∠BAD ∠BAD = 1/2∠ BOD
Similarly∠BCD = 1/2 ∠DOB ∠BAD + ∠BCD
= 1/2∠BOD + 1/2 ∠DOB
=1/2(∠ BOD + ∠DOB)
= (1/2)X360° = 180°
Similarly ∠B + ∠D = 180°
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