Question

${\large{\red{{\underline{\boxed{\bf{Question :}}}}}}}$
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.


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Answer 1

⇝Given :-

• A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m.

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⇝To Find :-

• Find the area of field.

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⇝According to Attachment :

${\red{\rightarrowtail}{\bf{AB \: ll \: CD}}}$

${\red{\rightarrowtail}{\bf{AB = 25m}}}$

${\red{\rightarrowtail}{\bf{CD = 10m}}}$

${\red{\rightarrowtail}{\bf{BC = 14m}}}$

${\red{\rightarrowtail}{\bf{AD = 13m}}}$

From point C draw CE ll DA .

Therefore, ABCE is a parallelogram having CD ll CE .

❒ Sides of front triangle :

${\red{\rightarrowtail}{\bf{ CB= 14m}}}$

${\red{\rightarrowtail}{\bf{CE = AD = 13m}}}$

${\red{\rightarrowtail}{\bf{BE = AB - AE = 25 - 10 = 15m}}}$

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⇝Formula Used :-

➵Area of triangle :

${\large{\red{\bigstar \: \: \: {\pink{\underbrace{\underline{\orange{\bf{Area = \sqrt{s(s - a)(s - b)(s - c)} }}}}}}}}}$

➵Area of parallelogram :

${\large{\red {\bigstar \: \: \: {\pink{\underbrace{\underline{\orange{\bf{Area = \frac{1}{2 } \times sum \: of \: ll \: sides\times h }}}}}}}}}$

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⇝Solution :-

❒ First Area of triangle :

✏ Semi - Perimeter :

${\large{:{\longmapsto{\bf{semi - perimeter = \frac{a + b + c}{2} }}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\large{:{\longmapsto{\bf{ \frac{14 + 13 + 15}{2} }}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\large{:{\longmapsto{\bf{{\cancel \frac{42}{2} }}}}}}$

${\orange{\large{:{\dashrightarrow{\underline{\blue{\bf{21 m}}}}}}}}$

✏ Area :

${\large{:{\longmapsto{\bf{Area = \sqrt{s(s - a)(s - b)(s - c)} }}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: {\large{:{\longmapsto{\bf{ \sqrt{21(21 - 14)(21- 13)(21 - 15)} }}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: {\large{:{\longmapsto{\bf{ \sqrt{21 \times 7 \times 8 \times 6} }}}}}$

${\orange{\large{:{\dashrightarrow{\underline{\blue{\bf{84 \: {m}^{2} }}}}}}}}$

✏ Area of triangle BEC :

${\large{:{\longmapsto{\bf{Area = \frac{1}{2} \times b \times h }}}}}$

${\large{:{\longmapsto{\bf{84 = \frac{1}{2} \times 15\times h }}}}}$

${\large{:{\longmapsto{\bf{h = {\cancel {84 \times ( \frac{2}{15}) } }}}}}}$

${\orange{\large{:{\dashrightarrow{\underline{\blue{\bf{H = 11.2 m}}}}}}}}$

✏ Area of trapezium :

${\large{:{\longmapsto{\bf{Area = \frac{1}{2} \times \: sum \: of \: ll \: sides \times \: h }}}}}$

${\large{:{\longmapsto{\bf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{2} \times (10 + 25)\times 11.2 }}}}}$

${\large{:{\longmapsto{\bf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{2} \times 35\times 11.2 }}}}}$

${\large{:{\longmapsto{\bf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{\cancel{2} } \times{\cancel{ 392 }}}}}}}$

${\large{\purple{:{\twoheadrightarrow{\underline{\overline{\boxed{\bf{Area = 196 m²}}}}}}}}}$

❒ Hence :

${\large{\underline{\red{\underline{\mathfrak{\pmb{Area = 196 m²}}}}}}}$

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