Area of Plane Geometrical Figures
Category : 9th Class
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
\[\text{=}\frac{1}{2}\times \text{base}\times \text{height}=\left( \frac{1}{2}\times \text{25}\times \text{1}0.\text{8} \right)\text{c}{{\text{m}}^{\text{2}}}=\text{135 c}{{\text{m}}^{\text{2}}}\]
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