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?

Steps of construction of a quadrilateral when four sides and one of its diagonal are given.
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.


$$\begin{gathered} area of square = area of quadrilateral ABCD \\ And area of quad ABCD = area ∆ABD +area∆BCD \\ area ∆ABD = \sqrt{s(s-a)(s-b)(s-c)} =\sqrt{6(6-3)(6-4)(6-5)}=\sqrt{6\times 3\times 2\times 1}=6 sq units \\ area ∆BCD = \sqrt{s(s-a)(s-b)(s-c)}=\sqrt{\frac{13}{2}\times (\frac{13}{2}-2)(\frac{13}{2}-5)(\frac{13}{2}-6)}=\sqrt{\frac{13}{2}\times \frac{9}{2}\times \frac{3}{2}\times \frac{1}{2}}=4.68 sq units \\ \therefore area of square = 6sq units +4.68 sq units = 10.68 sq units \\ Hence side of square = \sqrt{10.68 sq units}=3.26 units \end{gathered}$$