Question

if the area of a rhombus with side X is x square by 2 what is the length of its longer diagonal​

Answer 1

Answer:

Length of the longer diagonal is

$\bold{\frac{(\sqrt{3}+1)x}{\sqrt{2}}}$

Step-by-step explanation:

Let 2l and 2m be the diagonals of the given rhombus with l > m

since the diagonals of rhombus intersecct at right angles, we have

$l^2+m^2=x^2$.....(1)

Given:

Area of the rhombus $=\frac{x^2}{2}$

$\implies\:\frac{1}{2}*d_1*d_2=\frac{x^2}{2}$

$\implies\:\frac{1}{2}*(2l)*(2m)=\frac{x^2}{2}$

$\implies\:2lm=\frac{x^2}{2}$....(2)

$(1)+(2)\implies\:(l+m)^2=\frac{3x^2}{2}$

$\implies\:l+m=\frac{\sqrt{3}x}{\sqrt{2}}$......(3)

$(1)-(2)\implies\:(l-m)^2=\frac{x^2}{2}$

$\implies\:l-m=\frac{x}{\sqrt2}$.....(4)

$(3)+(4)\implies\:2l=\frac{\sqrt{3}x}{\sqrt{2}}+\frac{x}{\sqrt2}$

$\implies\:2l=\frac{\sqrt{3}x+x}{\sqrt{2}}$

$\implies\:2l=\frac{(\sqrt{3}+1)x}{\sqrt{2}}$

$\therefore$Length of the longer diagonal is

$\frac{(\sqrt{3}+1)x}{\sqrt{2}}$

Answer 2

Answer:

$\sqrt{3x}$