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9th Class Mathematics Mensuration Quadrilateral

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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