Question 11 MarkThe area of a square is 16 sq cm. Its perimeter is _______________ [16 cm/8 cm].Answer16 cmView full question & answer→
Question 21 MarkThe area of a square is (a × a) sq. cm. Its perimeter is _______________ [4a cm/8a cm].Answer4a cmView full question & answer→
Question 31 MarkPerimeter of a square of side 1 cm = _______________Answer4 cmView full question & answer→
Question 41 MarkSide of a regular triangle = _______________ $\left[\left(\frac{\text { Perimeter }}{3}\right) /(\right.$ Perimeter $\left.\times 3)\right]$Answer$\frac{\text { Perimeter }}{3}$View full question & answer→
Question 51 MarkBreadth of a rectangle = _______________ $\left[\left(\frac{\text { Perimeter }}{2}-\right.\right.$ Length $) /\left(\frac{\text { Perimeter }}{2} \div\right.$ Length $\left.)\right]$Answer$\frac{\text { Perimeter }}{2}-$ LengthView full question & answer→
Question 61 MarkLength of a rectangle = _______________ $\left[\left(\frac{\text { Perimeter }}{2}-\right.\right.$ Breadth $) /\left(\frac{\text { Perimeter }}{2} \div\right.$ Breadth $\left.)\right]$Answer$\frac{\text { Perimeter }}{2} -$ BreadthView full question & answer→
Question 71 MarkSide of a square = _______________ [(Perimeter ÷ 4)/(Perimeter – 4)]AnswerPerimeter $\div$ 4View full question & answer→
Question 81 MarkThe perimeter of an equilateral triangle = _______________Answer3 x SideView full question & answer→
Question 91 MarkThe perimeter of a regular hexagon = _______________Answer6 x SideView full question & answer→
Question 101 MarkThe perimeter of a regular triangle is _______________ [(3 × Side)/(3 + Side)]Answer3 x SideView full question & answer→
Question 111 MarkThe perimeter of a rectangle is _______________ [2(Length × Breadth)/2(Length + Breadth)]Answer2(Length + Breadth)View full question & answer→
Question 121 MarkThe perimeter of square is _______________ [(4 × Side)/(4 + Side)]Answer4 x SideView full question & answer→
Question 131 MarkThe length of the boundary of a figure is called its _______________ [Perimeter/Area]AnswerPerimeterView full question & answer→