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  • The diagonals of a rhombus are along the line <img alt="\mathrm{y = x \; and\; y = -x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7By%20%3D%20x%20%5C%3B%20and%5C%3B%20y%20%3D%2

The diagonals of a rhombus are along the line \mathrm{y = x \; and\; y = -x} if one of the vertex be (1, 1) and area of the rhombus be \sqrt{2} sq. unit. One of the other three vertices is:

Option: 1

(-1,1)


Option: 2

\mathrm{\frac{1}{2\sqrt{2}},\frac{1}{2\sqrt{2}}}


Option: 3

(-1,-1)


Option: 4

(-\sqrt{2},-\sqrt{2})


Answers (1)

best_answer

Let vertex A be (1, 1) clearly vertex C will be (–1, –1)

Let co-ordinates of B be (h, –h) then co-ordinates of D will be (–h, h) so that

\begin{aligned} & \mathrm{\mathrm{BD}=\sqrt{(2 h)^2+(-2 h)^2}} \\ \\& \mathrm{=\sqrt{4 h^2+4 h^2}=|2 \sqrt{2 h}|} \end{aligned}

Now Area of rhombus \mathrm{=\frac{1}{2} \mathrm{AC} \cdot \mathrm{BD}=\sqrt{2}}

\begin{aligned} & \Rightarrow \frac{1}{2}(2 \sqrt{2})(2 \sqrt{2} \mathrm{~h})=\sqrt{2} \\ \\& \Rightarrow \mathrm{h}=\frac{1}{2 \sqrt{2}} \end{aligned}

Hence vertices are \mathrm{B\left(\frac{1}{2 \sqrt{2}},-\frac{1}{2 \sqrt{2}}\right), C(-1,-1), D\left(-\frac{1}{2 \sqrt{2}}, \frac{1}{2 \sqrt{2}}\right)}

Posted by

Shailly goel

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