Draw a quadrilateral of sides 2cm ,3cm,4cm,6cm and diagonal 5cm. Draw a square with area equal to that of the quadrilateral .*H* *ow can we construct such a figure?*

Let AB = 3 cm , AD = 4 cm , BC = 2 cm,CD = 6 cm

We have choosen the sides in such a way that the triangle law of inequality holds .

Step 1 Draw a line BD = 5 cm (length of the diagonal)

Step 2 Draw an arc with centre B and radius = 3 cm .

Step 3 .Draw an arc with centre C and radius = 4 cm meeting previous arc at A.

Step 4 .Similarly on the opposite side of the diagonal,draw arcs with centre at B and D having radius 2 cm and 6 cm respectively.

Let these meet at C.

Hence ABCD is the required quadrilateral.

$areaofsquare=areaofquadrilateralABCD\phantom{\rule{0ex}{0ex}}AndareaofquadABCD=area\u2206ABD+area\u2206BCD\phantom{\rule{0ex}{0ex}}area\u2206ABD=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{6(6-3)(6-4)(6-5)}=\sqrt{6\times 3\times 2\times 1}=6squnits\phantom{\rule{0ex}{0ex}}area\u2206BCD=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{\frac{13}{2}\times (\frac{13}{2}-2)(\frac{13}{2}-5)(\frac{13}{2}-6)}=\sqrt{\frac{13}{2}\times \frac{9}{2}\times \frac{3}{2}\times \frac{1}{2}}=4.68squnits\phantom{\rule{0ex}{0ex}}\therefore areaofsquare=6squnits+4.68squnits=10.68squnits\phantom{\rule{0ex}{0ex}}Hencesideofsquare=\sqrt{10.68squnits}=3.26units$

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