Question

a field in the shape of a trapezium whose parallel sides are 25cm and 10cm. The non parallel sides are 14m and 13m. Find the area of the field ​

Answer 1

Answer:

hope you can find ur answer......

take help from it....

Step-by-step explanation:

2(25-10)=15(2)=*

2*=15

*=15/2

*=7.5

case -2

1/2*7.5*(25+10)

1/2*7.5*35

Answer 2

$\Large{\underline{\underline{\mathfrak{\red{Given:-}}}}}$

$\rm{A\: field\: in\:the \:shape\: of\:a\:trapezium whose\: parallel \:sides\: are \:25cm \:and\: 10cm.\:The \:non \:parallel \:sides \:are \:14m\: and\: 13m. }$

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$\Large{\underline{\underline{\mathfrak{\blue{To \: Find:-}}}}}$

$\rm{Area\: of \: the\: field.}$

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$\Large{\underline{\underline{\mathfrak{\pink{Formula\: used:-}}}}}$

• $\rm{Heron's \: Formula= \sqrt{s(s - a)(s - b)(s - c)} }$
• $\rm{Area \: of \: parallelogram=Base×Height}$

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$\Large{\underline{\underline{\mathfrak{\purple{Solution:-}}}}}$

$\rm{From \: C,\: draw \: CE \: ||\: DA.}$

$\rm{Clearly, \: ASCE \: is \: a \: parallelogram\: having\: AD||CE\: and \: DC||AE\: such \: that \: AS =13 \: m \: and DC \: 10m}$

∴$\rm{AE=DC=10m \: and \: CE=AD=13m}$

$\rm{BE=AB-AE=\: (25-10)m=15m}$

$\rm{Thus,\: in \: BCE, \: we \: have}$

$\rm{BC=14m, \: CF = 13 m \: and \: BE =15 m}$

$\rm{Let\: s \: be \: the\: semi\: perimeter\:of \: triangle\: BCE,\: Then,}$

$\rm{2s=BC+CE+BE=14+13+15=42}$

$\rm{→s=21}$

∴$\rm{Area\: of \: traingle\:BCE = \sqrt{21 \times (21 - 14) \times (21 - 13) \times (21 - 15)} }$

$\rm{→Area\: of \: traingle\: BCE= \sqrt{21 \times 7 \times 8 \times 6} }$

$\rm{→Area\: of \: traingle\:BCE= \sqrt{ {7}^{2} \times {3}^{2} \times {4}^{2} }=84 {m}^{2} }$

$\rm{Also,\:Area\: of \: traingle\: BCE = \frac{1}{2} (BE×CL)}$

$\rm{→84=\frac{1}{2}×15×CL}$

$\rm{→CL=\frac{168}{15}=\frac{56}{5}}$

$\rm{→Height\: of \: parallelogram\: ADCE=CL=\:\frac{56}{5}m}$

$\rm{∴Area\: of \: parallelogram \: ADCE = Base×Height\:=AE×CL=10×\frac{56}{5}=112{m}^{2}}$

$\rm{Hence,\: Area\: of \: trapezium\: ABCD= \: Area \: of \: parallelogram\: ADCE\: + \: Area \: of \: traingle\:BCE=(112+84){m}^{2}=196{m}^{2}}$

${\rm{\underline{\boxed{\boxed{\mathfrak{\purple{Area\: of \: trapezium\: ABCD=196{m}^{2}}}}}}}}$

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$\Large{\underline{\underline{\mathfrak{\green{Thanks}}}}}$