Question

a filed is in the shape of a trapezium. The parallel sides are 10m and 25m the non parallel sides are 14m and 13m. Find the area of the field. [ch: heron's formula]

Answer 1

GivEn:

• Parallel sides of a trapezium shaped field are 10 m and 25 m.

• Non - parallel sides of a trapezium shaped field are 14 m and 13 m.

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To find:

• Area of the field.

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SoluTion:

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ABCD is a trapezium.

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$\bf Where \begin{cases} & \text{AB || CD } \\ & \text{AB = 25 m} \\ & \text{CD = 10 m} \\ & \text{AD = 13 m } \\ & \text{BC = 14 m}\end{cases}$

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Now, Construct a line CE which is parallel to AD.

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Therefore,

• CE = AD = 13 m

• AE = CD = 10 m

• EB = AB - AE = (25 m - 10 m) = 15 m

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${\underline{\sf{\bigstar\;Now,\;In\; \triangle\;EBC}}}$

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we have,

• EC = 13 m

• BC = 14 m

• EB = 15 m

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Now, Using Heron's Formula,

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we have to find area of ∆ EBC -

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$\star\;{\boxed{\sf{\purple{Heron's\;Formula = \sqrt{s(s - a)(s - b)(s - c)}}}}}$

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where,

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s = semi - perimeter

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$\star\;\sf s = \dfrac{a + b + c}{2}$

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$\;\;\;\;\small\;\sf \underline{Putting\;values\;:}$

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$:\implies\sf s = \dfrac{13 + 14 + 15}{2}$

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$:\implies\sf s = \cancel{ \dfrac{42}{2}}$

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$:\implies\bf \red{s = 21\;m}$

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Now,

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$\star\;\sf Area_{\;\triangle\:EBC} = \sqrt{s(s - a)(s - b)(s - c)}$

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$\;\;\;\;\small\;\sf \underline{Putting\;values\;:}$

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$:\implies\sf \sqrt{21(21 - 13)(21 - 14)(21 - 15)}$

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$:\implies\sf \sqrt{21 \times 8 \times 7 \times 6}$

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$:\implies\sf \sqrt{7056}$

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$:\implies{\underline{\boxed{\bf{\pink{Area_{\;\triangle\:EBC} = 84\;m^2}}}}}\;\bigstar$

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Again, Area of ∆ EBC

using base and height,

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$\star\;{\boxed{\sf{\purple{Area = \dfrac{1}{2} \times B \times H}}}}$

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$\;\;\;\;\small\;\sf \underline{Putting\;values\;:}$

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$:\implies\sf 84 = \dfrac{1}{2} \times 15 \times H$

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$:\implies\sf H = 84 \times \dfrac{2}{15}$

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$:\implies\sf H = \cancel{ \dfrac{168}{15}}$

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$:\implies{\underline{\boxed{\bf{\blue{Height = \dfrac{56}{5}\;m}}}}}\;\bigstar$

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${\underline{\sf{\bigstar\;Area\;of\; trapezium\;:}}}$

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As we know that,

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$\star{\boxed{\sf{\purple{Area\;or\; trapezium = \dfrac{1}{2} \times (sum\;of\;||\;sides) \times H}}}}$

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$\;\;\;\;\small\;\sf \underline{Putting\;values\;:}$

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$:\implies\sf \dfrac{1}{2} \times (10 + 25) \times \dfrac{56}{5}$

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$:\implies\sf \dfrac{1}{ \cancel{2}} \times \cancel{35} \times \dfrac{ \cancel{56}}{ \cancel{5}}$

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$:\implies\sf 7 \times 28$

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$:\implies{\underline{\boxed{\bf{\green{Area_{\;(ABCD)} = 196\;m^2}}}}}\;\bigstar$

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$\therefore$ Hence, Area of trapezium shaped field is 196 m².