Videos Related with Area of Triangle and Quadrilateral

Note on Area of Triangle and Quadrilateral

  • Note
  • Things to remember
  • Videos
  • Exercise
  • Quiz

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,

fzsdf

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*}

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

 

 

xfg

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*}

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

 

c v

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*}

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

 

fvdv

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*}

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

 

xzc

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*}

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

  • Square, rectangle and rhombus are all parallelograms.
  • Kite and Trapezium are not parallelograms.
  • A square is a rhombus as well as a rectangle.
  • A parallelogram is a trapezium.

 

.

Very Short Questions

Solution:

SQ = \(3 \sqrt{2}\) cm
Let, PO = SP = x cm
In right angle \( \Delta PQS, \)

\begin{align*} SQ^2 \  &= PQ^2 + SP^2\\ or,(3 \sqrt{2})^2 &= x^2+ x^2\\ or, x^2 &= \frac{18}{2}\\ or,x^2&=9\\ \therefore x &=3cm \end{align*}

\begin{align*}Area \: of \: \Delta PQS &= \frac{1}{2}\times b \times h \\ &= \frac{1}{2}\times 3 \times 3\ cm^2 \\&=4.5\: cm^2 \end{align*}

\begin{align*}Area \: of \: \Delta PQT &= Area \: of \: \Delta PQT = 4.5 \: cm^2 \end{align*}

 

Solution:

QS = 6 cm, 2PA = 6cm \( \therefore PA = \frac{6}{2} = 3cm\)

3RB = 6cm \(\therefore RB = \frac{6}{3} = 2cm \)

Area of quad. PQRS = ?

\begin{align*} Area \: of \: PQRS &= \frac{1}{2} QS (PA + RB)\\ &=\frac{1}{2}\times 6 (3+2)\\ &= 3 \times 5\\ &= 15 \: cm^2 \: _{ans} \end{align*}

Solution:

BD = \(3 \sqrt{2}cm\)
BC = CD = AD = AB [\(\therefore\) Sides of square]

In right angled \(\Delta\)BCD

\begin{align*} (\sqrt{BD} \ )^2 &= (\sqrt{BC} \ )^2 +(\sqrt{CD} \ )^2\\ 3\sqrt{2} &= (\sqrt{BC} \ )^2 +(\sqrt{CD} \ )^2\\ 18 &= 2BC^2\\ BC^2 &= \frac{18}{2} \\ &= 9 \\ &= 3^2\: cm \end{align*}

Area of ABCD = 9 cm2

Solution:

PR = 10 cm
SM = 3.4 cm
QN = 4.6 cm

We know,

\begin{align*} Area \: of \: quad. PQRS &= \frac{1}{2}PR (SM + QN)\\&=\frac{1}{2}\times 10 (3.4+4.6)\\&=5\times8\\&=40 \:cm^2\:\:\: _{ans} \end{align*}

Solution:
BD = 12 cm
\(2CN = 12 cm\\ CN = \frac{12}{2} = 6\:cm \)
\(4AM = 12 cm\\ AM = \frac{12}{4} = 3\:cm \)
Area of quad. ABCD = ?
We know,
\begin{align*}Area\:of\:quad.ABCD &= \frac{1}{2}BD(AM+CN)\\ &=\frac{1}{2}\times12(3+6)\\&= 6 \times 9 \\ 6 \times 9\\ &= 54\:cm^2 \end{align*}

Solution:
BD = 8 cm
AN = 3 cm
CM = 6 cm
Area of quad. ABCD = ?
We know,
\begin{align*}Area\:of\:quad.ABCD &= \frac{1}{2}BD(AN+CM)\\ &=\frac{1}{2}\times8(3+6)\\&= 4 \times 9 \\ &= 36\:cm^2 \end{align*}

Soution:

WZ =\(2 \sqrt{2}\:cm\)
WX = XY = YZ = WZ [\(\therefore\) Sides of squares ]

In right angled \(\Delta \)WXY.

\begin{align*} (WX)^2 + (XY)^2 &=(WY)^2\\ (WX)^2 + (WX)^2 &=( 2 \sqrt{2} )^2\\ 2(WX)^2&=8\\(WY)^2&=\frac{8}{2}\\WY &=\sqrt{4}\\WY&=2\end{align*}

Now,

\begin{align*}\text{Area of square WXYZ} &= (WX)^2\\&=(2)^2 \\&=4\:cm^2 \:\:\:_{ans} \end{align*}

Solution:

QR = 10cm
RT = 6cm
Area of parallelogram, PQRS = ?

\begin{align*} PQRS &= base \times height \\&=QR \times RT \\&=10\times 6 \\&=60\:\:cm^2\:\:\:\:_{ans} \end{align*}

Solution:

PM= 10cm
PS= 5.7
QR= 6.3cm
Area of quad. PQRS = ?

\begin{align*} \text{Area of quad. PQRS} &= \frac{1}{2}h(a+b)\\ &= \frac{1}{2}\times 10(5.7+6.3)\\ &= 5 \times 12\\ &=60cm^2 \:\:\:\:_{}ans \end{align*}

Solution:

Area of parallelogram = \(56\:cm^2\)
Height of parallelogram = 7 cm
Base of parallelogram = ?

\begin{align*} \text{Area of parallelogram} &= base \times height\\ 56&=base \times 7\\base &= \frac{56}{7}\\&= 8 \: cm \end{align*}

Solution:

PQ = QR = RS = PS = 6cm \([\because sides\:of\: rhombus ]\)
PT = 4 cm
Area of rhombus PQRS = ?

We know,

\begin{align*} Area \: of \: Rhombus\: PQRS &= base \times height\\ &= QR \times PT\\ &= 6 \times 4\\ &= 24\:cm^2 \:\: _{Ans} \end{align*}

Solution:

NR = 8 cm
PM = 10 cm
MN = ?
Area of the MNRP = 72 \(cm^2\)
MA= NR = 8 cm
In the right angled triangle PMA,

\begin{align*} PA &= \sqrt{PM^2 -MA^2}\\&= \sqrt{10^2 - 8^2}\\ &=\sqrt{100-64}\\ &= \sqrt{36}\\ &= 6 cm \end{align*}

\begin{align*} \text{Area of quadrilateral MNRP} &= \frac{1}{2}h(a+b)\\ 72 &= \frac{1}{2}\times MA (MN + PR)\\ or, 72 &= \frac{1}{2}\times 8 (MN + PA+AR)\\ or, 72 &= 4(MN + 6 + MN)\\ or, \frac{72}{4}&= 2MN +6 \\ or, 18-6&=2MN\\ MN &= \frac{12}{2}\\ \therefore MN &= 6 cm\:\; _{ans} \end{align*}

Solution:

PR = 4cm
QS = 6 cm
Area of rhombus PQRS = ?

\begin{align*} \text{Area of rhombus PQRS} &= \frac{1}{2}d_1 \times d_2 \\ &= \frac{1}{2}\times PR \times QS\\ &= \frac{1}{2} \times 4 \times 6\\ &= 12 \: cm^2 \end{align*}

Solution:

Area of rhombus = 24 \(cm^2\)
Length of the diagonal \( d_1= 6 \: m\)
Length of other diagnal \(d_2 = ? \)
We know,
\begin{align*} Area \: of\: rhombus &= \frac{1}{2} \times d_1 \times d_2\\ or, 24 &= \frac{1}{2} \times 6 \times d_2 \\ or, 24 &= 3 \times d_2\\ or, d_2 &= \frac{24}{3}\\ \therefore d_2 &= 8\: cm \end{align*}

Solution:

Area of the \(\Delta KMN = 30\: cm^2\)
\(Area \: of\: the\: \Delta KMV = \frac {1}{2} \Delta KLN \\ [\because Median \: KM\: bisect\: the \: \Delta KLM ]\\ Area \: of \: \Delta KMN = \frac{1}{2}\times 30 = 15 \: cm^2 \:\:_{Ans} \)

Solution:

\( Area \: of \: \Delta ADE = 5 \: cm^2 \\ Area\: of\: \Delta ACE = Area \: of \: \Delta ADE = 5 \: cm^2 [\because The \: median \: AE \: bisect\: the \: \Delta ACD]\\ Area \: of \: \Delta ABC = Area \: of\: \Delta ACD \: [\because \text{The diagonal AC bisect the parallelogram ABCD}] \\ Area \: of \: \Delta ABC = \Delta ADE + \Delta ACE = 5+5 = 10\: cm^2 \\ The \: quad. ABCE = \Delta ABC + \Delta ACE = 10+ 5 = 15 \: cm^2 \:\: _{Ans.} \)

Solution:

AB = 12 cm
CD = 21 cm
MC = 15 cm
AD = 10 cm
\begin{align*} Here , DM &= CD - MC \\&= 21cm - 15cm \\ &= 6 cm \end{align*}

\(In \: Right \: Angled \: \Delta ADM, \)

\begin{align*} AM &= \sqrt{(AD)^2 - (DM^2)} \\ &= \sqrt{10^2 - 6^2 }\\ &= \sqrt{100 - 36}\\ &= \sqrt{64} \\ &= 8 cm \end{align*}

\begin{align*} Area \: of \: quadrilateral\:\: ABCD &= \frac{1}{2} \times h (a+b ) \\ &= \frac{1}{2} AM\:(AB +CD)\\ &= \frac{1}{2} \times 8(12+21)\\ &= 4 \times 33 \\ &= 132\:cm^2 \:\: _{Ans} \end{align*}

Solution:

DE = 12 cm
CE = 13 cm
AB = BC = CD = AD [\(\because\) Sides of square]

In right angled \(\Delta CDE, \)

\begin{align*} CD &= \sqrt {(CE)^2 - (DE)^2} \\ &= \sqrt{13^2 -12^2 }\\ &= \sqrt{169-144}\\ &= \sqrt{25} \\&=5 \:cm \\ \\ AE &= AD+DE\\ &= 5 +12\\ &= 17 \: cm \\ BC &= 5 cm \end{align*}

\begin{align*} Area \: of\: Trap. ABCE &= \frac{1}{2} h(a+b)\\&= \frac{1}{2} \times CD \:(AE + BC ) \\ &= \frac{1}{2} \times 5 (17 + 5)\\ &= 2.5 \times 22 \\ &= 55 \: cm^2 \: _{Ans} \end{align*}

Solution:

AB = 4 cm
Area of EBCF = ?

\begin{align*} Area \: of \: square ABCD &= (AB)^2 \\ &= 4^2 \\ &= 16 \: cm^2 \\ \\ Area \: of \: parallelogram \: EBCF &= Area \: of \: square \: ABCD \\ &= 16 \: cm^2 \:\: [\because \text{Standing n same base and between same parallel lines}] \end{align*}

Solution:

Area of \(\Delta PTQ = 6 \: cm^2\)
Area of quadrilateral QTSR = ?

Area of \(\Delta QTS \) =Area of \(\Delta PTQ = 6 \: cm^2[\because \text{ Median line QT bisect the} \: \Delta PQS]\)
\begin{align*}Area \: of\: \Delta QRS &= Area \:of \: \Delta PQT +Area \:of\: \Delta QTS \\&= 6 + 6 \\ &= 12 \: cm ^2 \end{align*}

\begin{align*}Area \: of \: quad \: QTSR &= Area \: of \: \Delta QTS + Area\: of\: \Delta QSR\\ &= 6 + 12 \\&= 18 \: cm^2 \end{align*}

Solution:

Area of \(\Delta MQR = 6.5 \: cm^2\)
Area of \(\Delta PQM = Area\: of\: \Delta MQR = 6.5 \: cm^2 \:\:[\because \text{Median line QM bisect the } \Delta PQR]\)
\begin{align*} Area \: of \: \Delta PRS &= Area \: of \: \Delta PRS\\ &= 6.5+6.5 \\ &= 13 \:cm^2\\ Area \: of \: \Delta PQS &= Area \: of \: \Delta PQR + area \: of\: \Delta PRS \\&= 13 + 13 \\ &= 26 \:\: cm^2 \:\:\: _{Ans.} \end{align*}

Solution:

QS = 6 cm
\(2PA = 6 cm\\ PA = \frac{6}{2}\\ \therefore PA = 3 cm\)
\(3RB = 6 cm\\ PA = \frac{6}{3}\\ \therefore PA = 2 cm\)

\begin{align*} Area \: of\: quadrilateral \: PQRS &= \frac{1}{2} QS (PA + RB) \\ &= \frac{1}{2} \times 6 (3 + 2 )\\ &= 3 \times 5 \\ &= 15 \: cm^2 \:\:\:\: _{Ans} \end{align*}

Solution:

MN = 10 cm
NT = 8 cm
MN \(\parallel\) BC and BC = 2MN \([\because \text{ M and N are midpoints of sides AB and AC}]\)
\(BC = 2 \times 10 = 20\:cm \)

\begin{align*} Area \: of \: quad. \: BMNC &= \frac{1}{2} h (a+b)\\ &= \frac{1}{2} \times NT (MN + BC)\\&= \frac{1}{2} \times 8 (10 + 20)\\ &= 4 \times 30 \\ &= 120 \: cm^2 \:\:\: _{Ans} \end{align*}

Solution:

BD = 8 cm
AN = 6 cm
CM = 3 cm
Area of ABCD = ?

\begin{align*} Area \: of\: ABCD &=\frac{1}{2} BD (AN + CM)\\ &= \frac{1}{2} \times 8 (6 + 3) \\ &= 4 \times 9 \\ &= 36 \: cm^2 \end{align*}

0%
  • PQRS is a quadrilateral in which PR = 10 cm, perpendicular from Sand Q on PR are 3.4 cm.and 4.6 respectively.Calculate the area of the quadrilateral.

    21 cm2


    30 cm2


    40 cm2


    35 cm2


  • The areas of two parallelograms are equal and their altitudes are 6cm and 9 c.If the base of the first parallelograms is 12 cm, find the base of the second parallelogram.

    9 cm


    2 cm


    5 cm


    8 cm


  • The area of the parallelograms are equal. The altitudes of one parallelogram is 4 cm and of the base of the other is 6 cm.If the base of the first parallelogram is 9 cm, find the height of the second  parallelogram.

    8 cm


    2 cm


    9 cm


    6 cm


  • You scored /3


    Take test again

DISCUSSIONS ABOUT THIS NOTE

You must login to reply

Forum Time Replies Report


You must login to reply

Hom pd Paudel

Area of parallelogram having diagonals 8cm and 6cm.


You must login to reply