Question

Length of the fence of a trapezium-shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.​

Answer 1

$\large\underline{ \underline{ \sf \maltese{ \: Question:- }}}$

• Length of the fence of a trapezium-shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

$\large\underline{ \underline{ \sf \maltese{ \: Given:- }}}$

• Length of Trapezium ABCD = 120m
• BC = 48m
• CD = 17m
• AD = 10m

$\large\underline{ \underline{ \sf \maltese{ \: To \: Find:- }}}$

• Area of the Trapezium Shaped Field .

$\large\underline{ \underline{ \sf \maltese{ \: Solution:- }}}$

$\begin{gathered}\begin{gathered}\begin{gathered}: \implies \underline\blue{ \boxed{\displaystyle \sf \bold\orange{\: AB \: + \:BC \: + \:CD \: + \: AD \: = \: 120m }} }\\ \\\end{gathered}\end{gathered}\end{gathered}$

$\qquad \quad {:} \longrightarrow \sf{\bf{ AB \: + \:BC \: + \:CD \: + \: AD \: = \: 120 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ AB \: + \:48\: + \:17 \: + \: 40 \: = \: 120 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ AB \: + \:105\: = \: 120 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ AB \: = \: 120 \: - \: 105 }}$

$\qquad \quad {:} \longrightarrow \sf{\bf{ AB \: = \: 15m }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{ \:AB\: = \: 15m }}}$

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Since AB is perpendicular to the parallel sides AD and BC . Therefore, AB is the altitude ( height ) of the trapezium.

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$\begin{gathered}\begin{gathered}\begin{gathered}: \implies \underline\blue{ \boxed{\displaystyle \sf \bold\orange{\: Area \: of \: Trapezium \: = \: \frac{1}{2} \: \times \: (Sum \: of \: Parallel \: sides \: ) \: \times \: Altitude }} }\\ \\\end{gathered}\end{gathered}\end{gathered}$

$\qquad \quad {:} \longrightarrow \sf{\bf{Area \: of \: Field \: = \: \frac{1}{2} \: \times \: (\: BC \: + \: AD \: ) \: \times \: AB }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Area \: of \: Field \: = \: \frac{1}{2} \: \times \: (\: 48\: + \: 40 \: ) \: \times \: 15 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Area \: of \: Field \: = \: \frac{1}{2} \: \times \: 88 \: \times \: 15 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Area \: of \: Field \: = \: \frac{1}{\cancel2} \: \times \: \cancel{88} \: \times \: 15 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Area \: of \: Field \: = \: 44 \: \times \: 15 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Area \: of \: Field \: = \: 660{m}^{2} }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{ \: Area \: of \: Field \: = \: 660{m}^{2} }}}$

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Answer 2

Answer:

Perimeter =120 m.

48+17+40+x=120

x=15 m.

Area =

2

1

(sum of parallel sides) × height

=

2

1

(48+40)×15

=660 m

2

.