## Quadrilateral

Category : 9th Class

### Quadrilateral

We know that a geometrical figure bounded by four lines segment is called quadrilateral. In this section we will study about area and perimetre of different quadrilaterals

Perimetre and Area of Rectangle

Let ABCD be a rectangle in which length $AB=\ell$ units, breadth BC = b units then we have:

(i) Area$=(l+b)$square units

(ii) $\text{Length}=\frac{area}{breadth},$$breadth=\frac{area}{length}$

(iii) Diagonal$=\sqrt{{{l}^{2}}+{{b}^{2}}}$units

(iv) Perimetre$=2(l+b)$units

Area of Walls of Room

Let I, b and h are length, breadth and height of the room respectively then area of walls of room$~=\{\text{2(}l+\text{b)}\times \text{h}\}$ sq units.

Perimetre and Area of Square

Let ABCD be a square with each side equal to 'a' units then we have (i) Area$={{a}^{2}}$sq units

(ii) Area$=\left\{ \frac{1}{2}{{\text{(diagonal)}}^{\text{2}}} \right\}$sq units

(iii)  Diagonal$=a\sqrt{2}$units

(iv)  Perimetre = 4a units

Perimetre and Area of Quadrilateral

(1)  When one diagonal and length of the perpendiculars from opposite vertices on it are given

$\therefore$ Let ABCD be a quadrilateral with diagonal AC

Let $BL\bot AC$& $DM\bot AC$

Then area of the quadrilateral with diagonal AC

= [(area of$\Delta ABC$) + (area of$\Delta ACD$)]sq. units

$=\left\{ \left( \frac{1}{2}\times AC\times BL \right)+\left( \frac{1}{2}\times AC\times DM \right) \right\}$sq units

$\therefore$area of the quadrilateral ABCD $=\left\{ \frac{1}{2}\times AC\times ({{h}_{1}}+{{h}_{2}}) \right\}$sq. units

where${{h}_{1}}$&${{h}_{2}}$ are the lengths of perpendiculars from opposite vertices to the diagonal AC.

(2)  When diagonals intersect at right angles

Let ABCD be a quadrilateral whose diagonals AC and BD intersect at O at right angles then area of the quadrilateral ABCD = [(area of$\Delta ABC$) + (area of$\Delta ACD$)] $=\left( \frac{1}{2}\times AC\times BO \right)+\left( \frac{1}{2}\times AC\times OD \right)$ $=\frac{1}{2}\times AC\times (BO+OD)$

$=\frac{1}{2}\times AC\times BD=\frac{1}{2}\times$ (product of diagonals)

Area of parallelogram = base x height

Let ABCD be a parallelogram and let $CL\bot AB$

Then area of parallelogram ABCD

$\text{=2}\times \text{(area of }\Delta \text{ABC})$ $\text{=}\left( \frac{1}{2}\times AB\times CL \right)\text{=(AB}\times \text{CL})=(\text{base}\times \text{height})$

similarly if $CM\bot AD$ then area of parallelogram$\text{ABCD}=(\text{AD}\times \text{CM})$$=(\text{base}\times$ $\text{bright})$

Area of a Rhombus

A parallelogram with all sides equal is called rhombus. The diagonal of a rhombus bisect each other at right angles. Area of rhombus $=\frac{1}{2}\times$ (product of diagonals)

Area of a Trapezium

Let ABCD be a trapezium in which $~\text{AB}\,\,\text{ }\!\!|\!\!\text{ }\,\,\text{ }\!\!|\!\!\text{ }\,\,\text{DC}$

& Let$CL\bot AB$. Draw$AM\bot CD$

Let $\text{CL}=\text{AM}=\text{h}$

Area of trapezium ABCD = [Area of$\Delta ABC$+ area of$\Delta ACD$]

$=\left[ \left( \frac{1}{2}\times AB\times h \right)\times \left( \frac{1}{2}\times CD\times h \right) \right]$

$=\left[ \frac{1}{2}(AB+CD)\times h \right]=\frac{1}{2}$(Sum of parallel sides)$\times$(distance among them)

The length of a rectangular plot of land is twice its breadth. If the perimetre of the plot be 20 m then find its area.

(a) $\text{245}0{{\text{m}}^{\text{2}}}$

(b) $\text{2251}{{\text{m}}^{\text{2}}}$

(c) $\text{556}0{{\text{m}}^{\text{3}}}$

(d)$\text{9}0\text{6}0{{\text{m}}^{\text{3}}}$

Answer: (a)

Explanation

Let the breadth of the plot be$x$meters then its length $=2x$meters $\therefore$its perimetre $=2\times$(length + breadth)

$=[2\times (2x+x)]\,\,metres=6x\,\,metres$

$\therefore$ $6x=210$ $\Rightarrow$ $x=35$

$\therefore$breadth = 35 m, length$=(\text{2}\times \text{35})\text{m}=\text{7}0\text{ m}$

$\therefore$ area of the plot = (length$\times$breath)

$=(\text{7}0\times \text{35}){{\text{m}}^{\text{2}}}=\text{245}0\text{ }{{\text{m}}^{\text{2}}}$

The longer side of a rectangular hall is 14 m and the length of its diagonal is 26 m. Find the area of the hall.

(a) 240m

(b) $\text{25}0{{\text{m}}^{\text{2}}}$

(c)$\text{24}0{{\text{m}}^{\text{2}}}$

(d) $\text{258}{{\text{m}}^{\text{2}}}$

(e) None of these

Answer: (c)

Explanation

Let ABCD be the hall in which AD = 24m & AC = 26 m

By Pythagoras theorem, we have:

$BC=\sqrt{A{{C}^{2}}-A{{B}^{2}}}$units$=\sqrt{{{(26)}^{2}}-{{(24)}^{2}}}$

$=\sqrt{(26-24)(26+24)}=\sqrt{50\times 2}=\sqrt{100}=10\,m$

Thus Length = 24 m, breadth = 10 m

$\therefore$area of the hall$=(\text{24}\times \text{1}0){{\text{m}}^{\text{2}}}=\text{24}0\text{ }{{\text{m}}^{\text{2}}}$

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