Math, asked by tarungautamqwerty, 11 months ago

pls answer this question.....Its maths question....

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Answered by annesha863

Hey mate I think that there is some missing of informations.

Although,

Area of a rhombus = 1/2×( Product of the diagonals)

∴Length of the smaller diagonal is = (2×area)/length of the greater diagonal.

tarungautamqwerty: hey mate i got answer just now in my notebook .....question is right....may i send you

annesha863: yes send

tarungautamqwerty: in rhombus

annesha863: it was given

annesha863: but, there is no ratio or length of the greater side

Answered by raziahmadmjg

$we \: knw \: that \: \\ area \: of \: rhmbus = \frac{1}{2} (product \: f \: diagonals) \\ area \: f \: rhombus = \\ \frac{1}{2} {x}^{2} + 2x + \frac{3}{2} \\ \frac{1}{2} ( {x}^{2} + 4x + 3) \\ \frac{1}{2} ( {x}^{2} + x + 3x + 3) \\ \frac{1}{2}x(x + 1) + 3(x + 1) \\ \frac{1}{2} (x + 1)(x + 3) \\$
Hence the small diagonal is (X+1)

tarungautamqwerty: its correct

annesha863: no mate you see carefully

annesha863: to get the area we multiply 1/2 with product of the diagonals

tarungautamqwerty: mine answer is also this. just do middle term splitting after equating area of rhombus and given equation

tarungautamqwerty: smaller diagnal is x+1 and larger is x+3

annesha863: so, here to get the length 2 should be multiplied

annesha863: ok

tarungautamqwerty: i hope it will clear your doubt

annesha863: if you are satisfied with this then do this

raziahmadmjg: small diagonal is x+1

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