Question

ABCD is a square and A (5, 7) and C (1, 10). Then find (i) the area of the square ABCD (ii) the perimeter of the square ABCD.​

Answer 1

`\mathrm{From\ figure,}\\\mathrm{AC=√((5-1)^2+(7-10)^2)=√(4^2+3^2)=5}\\\mathrm{Since\ all\ sides\ of\ squares\ are\ equal,}\\\mathrm{AB=BC=CD=AD}\\\mathrm{In\ right\ angled\ triangle\ ABC,}\\\mathrm{AC^2=AB^2+BC^2}\\\mathrm{or,\ 5^2=AB^2+AB^2}\\\mathrm{or,\ 25=2AB^2}\\\mathrm{or,\ AB^2=(25)/(2)}\\\\\mathrm{or,\ AB=(5)/(\sqrt2)\ units}`

`\mathrm{i.\ Solution:}\\\mathrm{Area\ of\ square=AB^2=(25)/(2)sq.\ units.}\\\mathrm{ii.\ Solution:}\\\mathrm{Perimeter\ of\ square=4AB=2((25)/(2))=50\ units.}`