Question

The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. Find the area of the trapezium.

Answer 1

Given:

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• Parallel sides of the trapezium = 25cm & 11cm
• Non-Parallel sides = 15cm & 13cm

To find:

• Area of trapezium?

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Solution:

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∴ Parallel sides are equal.

So,

• AB - DC = EB, EB = 15cm
• DA ≈ CE, ED = 14cm

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$\setlength{\unitlength}{1.2 cm}\begin{picture}(0,0)\linethickness{0.5mm}\qbezier(0,0)(0,0)(1,3)\qbezier(5,0)(5,0)(4,3)\qbezier(1,3)(1,3)(4,3)\qbezier(3,0)(8.2,0)(0,0)\put(-0.5,-0.3){ \sf A }\put(-0.5,-0.3){ \sf \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E}\put(5.3,-0.3){ \sf B }\put(4.2,3.1){ \sf C }\put(0.6,3.1){ \sf D }\end{picture}$

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We know that,

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$\begin{gathered}\begin{gathered}\star\;{\boxed{\sf{\purple{{Heron's \: formula } = \dfrac{a + b + c}{2}}}}} \\ \end{gathered}\end{gathered}$

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In ∆BDC,

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$\begin{gathered}\begin{gathered}:\implies\sf \dfrac{13 + 14 + 15}{2}\\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}:\implies\sf \dfrac{42}{2}\\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}:\implies{ {\boxed{\sf{\pink{ \sf 21}}}}} \: \: \bigstar \: \\ \\\end{gathered}\end{gathered}$

⠀━━━━━━━━━━━━━━━━━━━━━━━━

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Now, By using the formula given below,

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$\begin{gathered}\begin{gathered}\star\;{\boxed{\sf{\purple{{{ Area = \sqrt{s(s - a)(s - b) (s - c) } }}}}}} \\ \end{gathered}\end{gathered}$

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

$\begin{gathered}\begin{gathered}:\implies\sf \sqrt{21(6)(7) (8) }\\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}:\implies\sf \sqrt{7056 }\\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}:\implies{\boxed{\sf{\purple{{84 \: {cm}^{2} }}}}}\\ \\\end{gathered}\end{gathered}$

$\sf {\underline{Thus, Area \: of \: \triangle BDC \: is \: { \frak{ \pmb{84 \: cm^2}}}, Respectively}}$

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⠀━━━━━━━━━━━━━━━━━━━━━━━━

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Now, Let's find height of the trapezium,

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$\begin{gathered}\begin{gathered}\star\;{\boxed{\sf{\purple{{{ Area = \dfrac{1}{2} \times b \times h }}}}}} \\ \end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}:\implies\sf h = \dfrac{84 \times 2}{14}\\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}:\implies\sf h = \dfrac{168}{14}\\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}:\implies \boxed{\sf \pink{h = 12\: cm}} \: \bigstar\\ \\\end{gathered}\end{gathered}$

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$\sf {\underline{Thus, Height \: of \: trapezium \: is \: { \frak{ \pmb{12 \: cm}}}, Respectively}}$

⠀━━━━━━━━━━━━━━━━━━━━━━━━

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Lastly, Let's find area of trapezium,

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$\begin{gathered}\begin{gathered}\star\;{\boxed{\sf{\purple{{{ Area = \dfrac{a + b}{2} \times h }}}}}} \\ \end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}:\implies\sf \frac{1}{2} \times (11 + 25) \times 12\\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}:\implies\sf \frac{1}{2} \times 36 \times 12\\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}:\implies\sf \boxed{ \sf \pink{216 \: {cm}^{2}}} \: \bigstar\\ \\\end{gathered}\end{gathered}$

$\sf {\underline{Therefore, Area \: of \: trapezium \: is \: { \frak{ \pmb{216 \: cm^2}}}, Respectively}}$

Answer 2

Given : Parallel sides of a trapezium are 25 cm and 11 cm.

To Find : Area of the Trapezium ?

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Solution : Let ABC is the Trapezium such than,

AB//CD, AB = 11 cm, CD = 25 cm

AD = 15 cm & BC = 13 cm

$~$

• $\leadsto$We draw BE lr DC & BF//AD

$~$

$~~~~~~~~~~{\sf:\implies{ABDF~is~a~//^{gm}}}$

$~~~~~~~~~~{\sf:\implies{BF~=~15~cm~\&~DF~=~11~cm}}$

$~~~~~~~~~~{\sf:\implies{FC~=~25~-~11~=~14~cm}}$

$~$

${\sf\therefore{Area ~of~\triangle~=~BCF~=~\dfrac{1}{\cancel{2}}}~(\cancel{14})~×~BE~=~7BE~~~~~~~~~~~~~~~~~~~\bigg\lgroup{Eqⁿ~(i)\bigg\rgroup}}$

$~$

$\underline{\frak{As ~we~ know that~:}}$

$~$

• $\boxed{\sf\pink{Heron's ~Formula~=~\dfrac{A~+~B~+~C}{2}}}$

$~$

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

$~~~~~~~~~~{\sf:\implies{S~=~\dfrac{1}{2}(13~+~14~+~15)}}$

$~~~~~~~~~~{\sf:\implies{S~=~\cancel\dfrac{42}{2}}}$${\sf{=~21~cm}}$

$~~~~~~~~~~:\implies{\underline{\boxed{\frak{\pink{S~=~21~cm}}}}}$★

$~$

___________________________________

• $\leadsto$Find Area of Triangle ?

$\underline{\frak{As~ we ~know~ that~:}}$

• $\boxed{\sf\pink{Area~ of~\triangle~=~\sqrt{s(s~-~a)(s~-~b)(s~-~c)}}}$

$~$

$~~~~~~~~~~{\sf:\implies{Area ~of~\triangle~BCF~=~\sqrt{21(21~-~13)(21~-~14)(21~-~15)}}}$

$~~~~~~~~~~{\sf:\implies{Area ~of~\triangle~BCF~=~\sqrt{21(8)(7)(6)}}}$

$~~~~~~~~~~{\sf:\implies{Area ~of~\triangle~=~\sqrt{7056}}}$

$~~~~~~~~~~:\implies{\underline{\boxed{\frak{\pink{Area ~of~\triangle~=~84~cm^2}}}}}$★

$~$

Therefore,

$\therefore\underline{\sf{Area~ of~\triangle~BCF~is~\bf{84~cm^2}}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\bigg\lgroup{\sf{Eqⁿ~(ii)\bigg\rgroup}}$

$~$

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• $\leadsto$ Find Height of Trapezium ?

$~~~~~~~~~~{\sf:\implies{7BE~=~84}}$

$~~~~~~~~~~{\sf:\implies{BE~=~\cancel\dfrac{84}{7}}}$${\sf{=~12~cm}}$

• ${\underline{\boxed{\frak{\pink{Height ~of~Trapezium~=~12~cm}}}}}$★

$~$

Therefore,

$\therefore\underline{\sf{Height ~of~ the ~Trapezium~ is~\bf{12~cm}}}$

$~$

____________________________________

• $\leadsto$ Find Area of Trapezium ?

$\underline{\frak{As~ we~ know~ that~:}}$

• $\boxed{\sf\pink{Area~=~\dfrac{a~+~b}{2}~×~h}}$★

$~$

$~~~~~~~~~~{\sf:\implies\therefore{Area~of~the~Trapezium~=~\dfrac{1}{\cancel{2}}(25~+~11)~×~\cancel{12}}}$

• $\underset{\blue {\rm Required\ Answer}}{\underbrace{\boxed{\frak{\pink{Area ~of ~the~ Trapezium~=~216~cm^2}}}}}$

$~$

Hence,

$\therefore\underline{\sf{Area ~of ~the ~Trapezium~is~\bf{216~cm^2}}}$

$~~~~$$\qquad\quad\therefore\underline{\textsf{\textbf{Hence Verified!}}}$