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Polygons(n>3)

All the three...

Question

All the three quadrillaterals I, II and III are squares. If the area of $1 = x^{2}$ and area of $I I = y^{2} (x^{2} > y^{2})$ . then the area of III is .......... .

2 x^{2}

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2 (x^{2} - y^{2})

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x^{2} + y^{2}

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x^{2} - y^{2}

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Solution

The correct option is D $x^{2} - y^{2}$
All the three quadrillaterals I, II and III are squares. If the area of 1=x2 and area of II=y2.
Then length of side( i )square= $\sqrt{x^{2}} = x$
And length of side (||) square= $\sqrt{y^{2}} = y$
So length of side (|||) = $\sqrt{x^{2} - y^{2}}$
Then area of (|||)= $x^{2} - y^{2}$

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