Area of Triangle and Quadrilateral

The triangle is a three-sided polygon and quadrilateral are four sided polygon. Since triangle and quadrilateral are both closed plane figures, they divide the plane into an interior and exterior regions. The notation of area of triangle and quadrilateral, used in daily life always means that measure of the extent of the interior region.

Area of a Triangle

Study the following figures,

In each figure, BC is the base (b) and AD is the height (h).

In figure 1

\begin{align*} Area \: of \Delta ABD &= \frac {1}{2} BD \times AD (\therefore AD ⊥ BD, \text{so AB is the height and BD is the base}) \\ Area \: of \Delta ADC &= \frac{1}{2} DC \times AD (\therefore AD ⊥ DC, \text{so AD is the height and DC is the base}) \\ \therefore Area \: of \Delta ABC &= Area \: of \Delta ABD + Area \: of \Delta ADC \\ &= \frac {1}{2} BD \times AD + \frac{1}{2} DC \times AD\\ &= \frac{1}{2} AD \: (BD + DC) \\ &= \frac {1}{2} h \times BC \\ &= \frac{1}{2} b \times h \: \: \: \: (\because BC = b ) \end{align*}

In figure 2

BC is the base and AB is the height. So as above,

$$ Area \: of \Delta ABC= \frac{1}{2} base \: (b) \times height \: (h)$$

In figure 3

\begin{align*} Area \: of \Delta ABC &= \frac {1}{2}DC \times AD \\Area \: of \Delta ADB &= \frac {1}{2} DB \times AD \\ \therefore Area \: of \Delta ABC &=Area \: of \Delta ABC -Area \: of \Delta ADB \\ &=\frac {1}{2}DC \times AD -\frac {1}{2} DB \times AD \\ &= \frac {1}{2} AD (DC - DB)\\ &= \frac{1}{2} AD \times BC\\ &= \frac{1}{2} \: b \times h \end{align*}

\( \therefore Area \: of \: a \: triangle = \frac {1}{2} base \times height \)

Area of Rhombus

Let ABCD be a rhombus where AC and BD are its diagonals. We know that diagonals of rhombus bisect each other at a right angle.
i.e AO = CO and BO = DO

\begin{align*} \therefore \text{Area of Rhombus ABCD} &= Area \: of \Delta ABO + Area \: of \Delta ADO + Area \: of \Delta BCO + Area \: of \Delta CDO \\ &=\frac{1}{2} \times AO \times BO + \frac{1}{2} \times DO \times AO + \frac{1}{2} \times BO \times CO + \frac{1}{2} \times DO \times CO\\ &= \frac{1}{2} \times AO \times (BO + DO) + \frac{1}{2} \times CO \times (BO + DO)\\ &= \frac{1}{2} \times AO \times BD + \frac{1}{2} \times CO \times BD\\ &= \frac{1}{2} \times BD \times (AO + CO)\\ &= \frac{1}{2} \times BD \times AC \\ &= \frac{1}{2} \times d_1 \times d_2 \: \: \: \: \: \: \: \: \: \: [Where \: d_1 = BD \: and \: d_2 = AC] \end{align*}

\( \boxed{\therefore \text{The area of a rhombus}= \text{one half of the product of its diagonals.}} \)

Area of a kite

Let ABCD be a kite with AC and BD as its diagonals. We know that diagonals of a kite intersect at right angles.

So,
\begin{align*} \text{Area of kite ABCD} &= Area \: of \Delta ABO +Area \: of \Delta ADO +Area \: of \Delta BCO +Area \: of \Delta CDO\\ &= \frac {1}{2} \times BO \times AO +\frac {1}{2} \times DO \times AO +\frac {1}{2} \times BO \times CO +\frac {1}{2} \times DO \times CO \\ &=\frac {1}{2} \times AO \times (BO + DO) +\frac {1}{2} \times CO \times (BO + DO)\\ &=\frac {1}{2} \times AO \times BD +\frac {1}{2} \times CO \times BD \\ &= \frac {1}{2} \times BD \times (AO + CO) \\ &= \frac{1}{2} \times BD \times AC \\ &= \frac{1}{2} \times d_1 \times d_2 \:\:\:\:\: \: [Where, \: BD= d_1 \: and \: AC=d_2] \\ \end{align*}

\( \boxed{\therefore \text {Area of a kite = One half of the product of its two diagonals}} \)

Area of Trapezium

Let ABCD is a trapezium in which AD // BC. AD and BC are its bases. Draw AE⊥ BC, where AE is the height.
Now, construct AF // DC then FC = AD = b (say) and let BC = b '. From the figure given above,

\begin{align*} \text {Area of trapezium ABCD} &= Area \: of \Delta ABF + Area \: of \: parallelogram AFCD \\ &= \frac{1}{2} \times BF \times AE \times FC \times AE \: \: \: [ \because \text {Area of parallelogram =} base \times height ] \\ &= \frac {1}{2} \times (BC - FC) \times AE \times FC \times AE \\ &= \frac {1}{2} \times BC \times AE - \frac{1}{2} \times FC \times AE + FC \times AE \\ &=\frac{1}{2} \times b' \times h - \frac {1}{2} \times b \times h + b \times h \\ &= \frac {1}{2} (b'h + bh)\\ &= \frac {1}{2} h(b + b')\\ \end{align*}

\( \boxed {\therefore \text{Area of a trapezium =} \frac{1}{2} \times height \times sum \: of \: the \: base } \)

Area of a quadrilateral

Let ABCD be a quadrilateral and BD be its diagonal.
Draw AM⊥ BD and CN⊥ BD.
Let AM = h₁, CN =h₂ and BD = d
Then,
\begin{align*} \text{Area of quadrilateral ABCD} &= Area \: of \Delta ABD + Area \: \Delta BCD \\ &= \frac {1}{2} \times BD \times AM + \frac {1}{2} \times BD \times CN \\ &= \frac {1}{2} \times d \times h_1 + \frac {1}{2} \times d \times h_2 \\ &= \frac{1}{2} d(h_1 + h_2) \end{align*}

\( \boxed {\therefore \text {Area of a quadrilateral = one half of a diagonal and sum of the perpendiculars on it from the opposite vertices } } \)