# 9th Class Mathematics Mensuration Area of Plane Geometrical Figures

Area of Plane Geometrical Figures

Category : 9th Class

### 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

Explanation

Area of the given triangle

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

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