Perimeter and Area of Plane Figures

 Triangle Let $a, b, c$$a, b, c$ denotes the sides of the Triangle. Then: $\displaystyle \text{Perimeter } = a + b + c$$\displaystyle \text{Perimeter } = a + b + c$ $\displaystyle \text{Semi-Perimeter }(s) = \frac{1}{2} (a + b + c)$$\displaystyle \text{Semi-Perimeter }(s) = \frac{1}{2} (a + b + c)$ $\displaystyle \text{Area } = \sqrt{s(s-a)(s-b)(s-c)}$$\displaystyle \text{Area } = \sqrt{s(s-a)(s-b)(s-c)}$ $\displaystyle \text{Area } = \frac{1}{2} \times Base \times Height = \frac{1}{2} bh$$\displaystyle \text{Area } = \frac{1}{2} \times Base \times Height = \frac{1}{2} bh$ Right-angled Triangle Let $b$$b$ be the base, $h$$h$ be the height (or perpendicular) and $a$$a$ be the Hypotenuse. Then, $\displaystyle \text{Perimeter } = a + b + h$$\displaystyle \text{Perimeter } = a + b + h$ $\displaystyle \text{Area } = \frac{1}{2} \times Base \times Height = \frac{1}{2} bh$$\displaystyle \text{Area } = \frac{1}{2} \times Base \times Height = \frac{1}{2} bh$ $\displaystyle \text{Hypotenuse }= \sqrt{b^2 + h^2}$$\displaystyle \text{Hypotenuse }= \sqrt{b^2 + h^2}$ Isosceles Right – angled triangle Let the equal sides be $a$$a$. $\displaystyle \text{Hypotenuse }= \sqrt{a^2+a^2} = \sqrt{2} a$$\displaystyle \text{Hypotenuse }= \sqrt{a^2+a^2} = \sqrt{2} a$ $\displaystyle \text{Perimeter } = 2a + \sqrt{2}a$$\displaystyle \text{Perimeter } = 2a + \sqrt{2}a$ $\displaystyle \text{Area } = \frac{1}{2} \times Base \times Height = \frac{1}{2} a^2$$\displaystyle \text{Area } = \frac{1}{2} \times Base \times Height = \frac{1}{2} a^2$ Equilateral Triangle Let each of the side is $a$$a$. Then $\displaystyle \text{Perimeter } = 3a$$\displaystyle \text{Perimeter } = 3a$ $\displaystyle \text{Height }= \frac{\sqrt{3}}{2} a$$\displaystyle \text{Height }= \frac{\sqrt{3}}{2} a$ $\displaystyle \text{Area } = \frac{1}{2} \times Base \times Height = \frac{1}{2} a \times \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{4} a^2$$\displaystyle \text{Area } = \frac{1}{2} \times Base \times Height = \frac{1}{2} a \times \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{4} a^2$ Isosceles Triangle Let the equal sides be $a$$a$. Let the base be $2b$$2b$. Then $\displaystyle \text{Perimeter } = 2a+ 2b$$\displaystyle \text{Perimeter } = 2a+ 2b$ $\displaystyle \text{Height }= \sqrt{a^2 - b^2}$$\displaystyle \text{Height }= \sqrt{a^2 - b^2}$ $\displaystyle \text{Area } = \frac{1}{2} \times 2b \times \sqrt{a^2 - b^2} = b \sqrt{a^2 - b^2}$$\displaystyle \text{Area } = \frac{1}{2} \times 2b \times \sqrt{a^2 - b^2} = b \sqrt{a^2 - b^2}$ Rectangle Let the length $= l$$= l$ and breadth $= b$$= b$. Then $\displaystyle \text{Perimeter } = 2 (l+b)$$\displaystyle \text{Perimeter } = 2 (l+b)$ $\displaystyle \text{Area } = lb$$\displaystyle \text{Area } = lb$ $\displaystyle \text{Diagonal }= \sqrt{l^2 + b^2}$$\displaystyle \text{Diagonal }= \sqrt{l^2 + b^2}$ Square Let the side of the square $= a$$= a$. Then $\displaystyle \text{Perimeter } = 4a$$\displaystyle \text{Perimeter } = 4a$ $\displaystyle \text{Area } = a^2$$\displaystyle \text{Area } = a^2$ $\displaystyle \text{Diagonal }= \sqrt{2} a$$\displaystyle \text{Diagonal }= \sqrt{2} a$ Parallelogram Let the two adjacent sides of the parallelogram be $a$$a$ and $b$$b$ $\displaystyle \text{Perimeter } = 2 (a+b)$$\displaystyle \text{Perimeter } = 2 (a+b)$ $\displaystyle \text{Area } = Base \times Height$$\displaystyle \text{Area } = Base \times Height$ Rhombus A parallelogram that has all the sides equal is called a rhombus. If $d_1$$d_1$ and $d_2$$d_2$ are the diagonals, then $\displaystyle \text{Side }= \frac{1}{2} \sqrt{{d_1}^2 + {d_2}^2}$$\displaystyle \text{Side }= \frac{1}{2} \sqrt{{d_1}^2 + {d_2}^2}$ $\displaystyle \text{Perimeter } = 4 \times Side = 2\sqrt{{d_1}^2 + {d_2}^2}$$\displaystyle \text{Perimeter } = 4 \times Side = 2\sqrt{{d_1}^2 + {d_2}^2}$ $\displaystyle \text{Area } = \frac{1}{2} d_1 d_2$$\displaystyle \text{Area } = \frac{1}{2} d_1 d_2$ Trapezium A trapezium is a quadrilateral two of whose sides are parallel. A trapezium whose non-parallel sides are equal is known as an isosceles trapezium. Let $a$$a$ and $b$$b$ be the parallel sides and $h$$h$ be the distance between the parallel sides. Then $\displaystyle \text{Area } = \frac{1}{2} (a+b) \times h$$\displaystyle \text{Area } = \frac{1}{2} (a+b) \times h$ Quadrilateral If $h_1$$h_1$ and $h_2$$h_2$ are the perpendicular distances from the diagonal $AC$$AC$ of the quadrilateral $ABCD$$ABCD$ from the vertices $B$$B$ and $D$$D$ respectively. Then, $\displaystyle \text{Area } = \frac{1}{2} (AC)(h_1 + h_2)$$\displaystyle \text{Area } = \frac{1}{2} (AC)(h_1 + h_2)$