Question

The perimeter of trapezium shown in Fig. 13.13 is 120 m. If BC = 48 m, CD = 17 m B
and AD = 40 m, find the area of this field. It is given that side AB is perpendicular to the
parallel sides AD and BC.
​

Answer 1

${\large{\underline{\underline{\rm{Required\;Solution:-}}}}}$

$\green{\bigstar}$ Consider the attachment as the figure of the given question.

We know,

• BC = 48 m
• CD = 17 m
• AD = 40 m
• Perimeter = 120 m

As, AB is perpendicular to BC, it will act as the height of the trapezium.

So, we have to find AB so as to find the area of the trapezium.

${\textit{\textbf{Finding\;AB}}}$

$\colon\rightarrow\sf \: Perimeter = Sum\:of\:all\:sides\\\\\sf\colon\rightarrow\sf\: Perimeter= AB+BC+CD+AD\\\\\sf\colon\rightarrow\:120 m = AB+48m+17m+40m \\\\\sf\colon\rightarrow\: 120m = AB+ 105m \\\\\sf \colon\rightarrow\:AB = 120m-105m\\\\\colon\rightarrow\boxed{\bf{AB=15m}}$

${\textit{\textbf{Finding\:Area}}}$

$\boxed{ \sf{area _{trapezium} = \frac{1}{2} (sum \: of \: parallel \: sides) \times height}}$

$\colon\implies\sf Area_{trapezium}=\dfrac{1}{2}(AD+BC)\times AB \\ \\\colon\implies\sf Area_{trapezium}=\dfrac{1}{2}(40+48)\times 15\:m^2\\ \\ \colon\implies\sf Area_{trapezium}=\dfrac{1}{\cancel2}\times \cancel{88}\times 15 \:m^2\\ \\ \colon\implies \sf Area_{trapezium}=44\times15 \:m^2\\ \\\colon\implies\underline{\boxed{\tt{\pink{Area_{trapezium}=\bf{\purple{660\:m^2}}}}}}$

So,

$\green{\bigstar}$ The area of the given trapezium would be 660 m² .

_____________________________

Answer 2

Given: The perimeter of trapezium is 120 m.

And, Length of sides BC, CD and AD is 48 m, 17 m and 40 m respectively.

And, Length of sides BC, CD and AD is 48 m, 17 m and 40 m respectively.To find: Area of trapezium field?

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀

❍ AB is perpendicular to the parallel sides AD and BC.

⠀⠀⠀⠀

$\begin{gathered}\dag\:{\underline{\frak{As\:we\:know\:that,\::}}}\\\\\end{gathered}$

$\begin{gathered}\star\:{\boxed{\sf{\pink{Perimeter_{\:(trapezium)} = a + b + c + d}}}}\\\\\end{gathered}$

Where,

a, b, c and d are the sides of trapezium.

⠀⠀⠀⠀

$\begin{gathered}\dag\:{\underline{\frak{Putting\:values\:in\:formula\::}}}\\\\\\ :\implies\sf AB + BC + CD + DA = 120\\\\\\ :\implies\sf AB + 48 + 17 + 40 = 120\\\\\\ :\implies\sf AB + 105 = 120\\\\\\ :\implies\sf AB = 120 - 105\\\\\\:\implies{\underline{\boxed{\frak{\purple{AB = 15\:m}}}}}\\\\\end{gathered}$

$\therefore\:{\underline{\sf{Height\:of\:trapezium\:field\:is\: {\textsf{\textbf{15\:m}}}.}}}$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀

▪︎Now, Finding Area of trapezium field,

⠀⠀⠀⠀

$\begin{gathered}\star\:{\boxed{\sf{\pink{Area_{\:(trapezium)} = \dfrac{1}{2} \times (a + b) \times h}}}}\\\\\end{gathered}$

Where,

a, b are two parallel sides and h is the height of distance between two parallel sides of trapezium.

⠀⠀⠀⠀

$\begin{gathered}\dag\:{\underline{\frak{Putting\:values\:in\:formula\::}}}\\\\\\ :\implies\sf Area_{\:(field)} = \dfrac{1}{2} \times (AD + BC) \times AB\\\\\\ :\implies\sf Area_{\:(field)} = \dfrac{1}{2} \times (40 + 48) \times 15\\\\\\ :\implies\sf Area_{\:(field)} = \dfrac{1}{\cancel{2}} \times \cancel{88} \times 15\\\\\\ :\implies\sf Area_{\:(field)} = 44 \times 15\\\\\\ :\implies{\underline{\boxed{\frak{\purple{Area_{\:(field)} = 660\:m^2}}}}}\\\\\end{gathered}$

$\therefore\:{\underline{\sf{Hence,\:Area\:of\:trapezium\:field\:is\: \bf{660\:m^2}.}}}$