draw a rhombus of 6cm and one angle 60? and draw right triangles of same area

Step - 1) Draw a line AB = 6cm.

Step - 2) At A, draw a line AQ so that angle (BAQ) = 60°.

Step - 3) Cut AQ at D such that AD = 6cm.

Step - 4) At B, draw a BP so that angle (ABP) = 180° - 60° i.e. 120°.

Step - 5) Cut BP at C such that BC = 6cm.

Step - 6) Join CD.

ABCD is the required rhombus.


$$\begin{gathered} \mathrm{Now} \mathrm{since} \mathrm{we} \mathrm{know} \mathrm{that} \mathrm{diagonal} \mathrm{bisct} \mathrm{angles} \mathrm{so} \mathrm{we} \mathrm{will} \mathrm{get} \mathrm{to} \mathrm{know} \mathrm{that} \\ \mathrm{it} \mathrm{is} \mathrm{a} \mathrm{joint} \mathrm{of} \mathrm{two} \mathrm{triangle} \mathrm{of} \mathrm{equilateral} \mathrm{each} \\ \mathrm{so} \mathrm{here} \mathrm{area} \mathrm{of} \mathrm{complete} \mathrm{rhombus} \\ =2\times \frac{\sqrt{3}}{4}\times {\mathrm{side}}^2= 2\times \frac{\sqrt{3}}{4}\times 36 \mathrm{square} \mathrm{cm} \\ =18\sqrt{3} {\mathrm{cm}}^2 \\ \mathrm{now} \mathrm{without} \mathrm{any} \mathrm{fixed} \mathrm{side} \mathrm{of} \mathrm{right} \mathrm{angle} \mathrm{triangle} \mathrm{we} \mathrm{cannot} \mathrm{make} \mathrm{confirmed} \\ \mathrm{triangle} \mathrm{as} \mathrm{there} \mathrm{can} \mathrm{be} \mathrm{infinite} \mathrm{right} \mathrm{triangle} \mathrm{with} \mathrm{this} \mathrm{area}. \end{gathered}$$