9th Class Mathematics Related to Competitive Exam Area of Plane Geometrical Figures

       Area of Plane Geometrical Figures  

            Area of Plane Figures

The area of a plane figure is the measurement of the surface enclosed by its boundry. In this chapter we will study about different plane figures with its area. 

Area of Triangle

Area of \[\Delta \text{ABC}=\frac{1}{2}\text{BC}\times \text{AD}\] square unit

Area of right triangle \[=\frac{1}{2}\times \text{(perpendicular)}\times \text{Base}\] \[=\frac{1}{2}\times AB\times \text{BC}\]

Area of Triangle by Heron's Formula

Let a, b, c be the length of sides of a triangle

then area \[=\sqrt{s(s-a)(s-b)(s-c)}\]sq. unit where \[s=-\frac{1}{2}(a+b+c)\]  

Area of Equilateral Triangle

Area \[=\frac{\sqrt{3}}{4}{{(side)}^{2}}=\frac{\sqrt{3}}{4}{{a}^{2}}\] Area of isosceles triangle

\[=\frac{1}{2}\times BC\times AD=\frac{1}{4}b\sqrt{4{{a}^{2}}-{{b}^{2}}}\]  

Find the area of the triangle whose base = 25 cm and height = 10.8 cm.

(a) \[\text{125c}{{\text{m}}^{\text{3}}}\]                                            

(b) \[\text{135c}{{\text{m}}^{\text{2}}}\]      

(c) \[\text{124c}{{\text{m}}^{\text{2}}}\]                                            

(d) \[\text{199}\,\text{c}{{\text{m}}^{\text{2}}}\]

(e) None of these  

Answer: (b)

Explanation

Area of the given triangle

\[\text{=}\frac{1}{2}\times \text{base}\times \text{height}=\left( \frac{1}{2}\times \text{25}\times \text{1}0.\text{8} \right)\text{c}{{\text{m}}^{\text{2}}}=\text{135 c}{{\text{m}}^{\text{2}}}\]