Question

find the area of a triangle field whose sides are 30m, 40m and 50m. also, find the length of ?​

Answer 1

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Correct Question:-

find the area of a triangle field whose sides are 30m, 40m and 50m. also, find the length of altitude drawn on the longest side..‽

GivEn:

Sides of ∆ = 30 m, 40 m and 50 m

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To find:

Length of altitude corresponding to the largest side of the ∆.

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Solution:

$\begin{gathered}\frak{Here} \begin{cases} & \sf{a = 30\;m} \\ & \sf{b = 40\;m} \\ & \sf{c = 50\;m} \end{cases}\\ \\\end{gathered}$

✇ Finding semi - perimeter of ∆,

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$\begin{gathered}:\implies\sf s = \dfrac{a + b + c}{2}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf s = \dfrac{30 + 40 + 50}{2}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf s = \dfrac{120}{2}\\ \\\end{gathered}$

$\begin{gathered}:\implies\bf s = 60\;m\\ \\\end{gathered}$

✇ Now, Finding area of triangular field using Heron's Formula,

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

$\begin{gathered}:\implies\sf \sqrt{60(60 - 30)(60 - 40)(60 - 50)}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \sqrt{60 \times 30 \times 20 \times 10}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \sqrt{360000}\\ \\\end{gathered}$

$\begin{gathered}:\implies{\boxed{\frak{\pink{600\;m^2}}}}\;\bigstar\\ \\\end{gathered}$

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✇ Now, Finding length of altitude corresponding to the largest side of the ∆.

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$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\linethickness{0.3mm}\qbezier(1, 0)(1,0)( 3,3)\qbezier(5,0)(5, 0)(3,3)\qbezier(5,0)(1,0)(1,0)\qbezier(1,0)(1,0)(4,1.5)\put(2.8,3.2){ \sf A }\put(2.5, - 0.3){40 m}}\put(0.5,-0.2){\sf B}\put(4.5,1){50 m}\put(5.1, - 0.3){\sf C}\put(1.1,1.8){30 m}\put(4.2, 1.5){\sf D}\end{picture}$

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Here,

Largest side of triangular field is = 50 m

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$\begin{gathered}\star\;{\boxed{\sf{\purple{Area = \dfrac{1}{2} \times (base) \times (height)}}}}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf 600 = \dfrac{1}{ \cancel{2}} \times \cancel{50} \times (height)\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf 600 = 25 \times (height)\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf Height = 600 \times \dfrac{1}{25}\\ \\\end{gathered}$

$\begin{gathered}:\implies{\boxed{\frak{\pink{Height = 24\;m}}}}\;\bigstar\\ \\\end{gathered}$

$\therefore\;{\underline{\sf{Hence,\;length\;of\; altitude\;is\; \bf{24\;m}.}}}$

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Answer 2

Answer:

GivEn:

Sides of ∆ = 30 m, 40 m and 50 m

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To find:

Length of altitude corresponding to the largest side of the ∆.

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Solution:

$\frak{Here} \begin{cases} & \sf{a = 30\;m} \\ & \sf{b = 40\;m} \\ & \sf{c = 50\;m} \end{cases}$

☞︎︎︎Finding semi - perimeter of ∆,

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$:\implies\sf s = \dfrac{a + b + c}{2}$

$:\implies\sf s = \dfrac{30 + 40 + 50}{2}$

$:\implies\sf s = \dfrac{120}{2}$

$:\implies\bf s = 60\;m$

☞︎︎︎Now, Finding area of triangular field using Heron's Formula,

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

$:\implies\sf \sqrt{60(60 - 30)(60 - 40)(60 - 50)}$

$:\implies\sf \sqrt{60 \times 30 \times 20 \times 10}$

$:\implies\sf \sqrt{360000}$

$:\implies{\boxed{\frak{\pink{600\;m^2}}}}\;\bigstar$

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☞︎︎︎ Now, Finding length of altitude corresponding to the largest side of the ∆.

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$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\linethickness{0.3mm}\qbezier(1, 0)(1,0)( 3,3)\qbezier(5,0)(5, 0)(3,3)\qbezier(5,0)(1,0)(1,0)\qbezier(1,0)(1,0)(4,1.5)\put(2.8,3.2){ \sf A }\put(2.5, - 0.3){40 m}}\put(0.5,-0.2){\sf B}\put(4.5,1){50 m}\put(5.1, - 0.3){\sf C}\put(1.1,1.8){30 m}\put(4.2, 1.5){\sf D}\end{picture}$

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Here,

Largest side of triangular field is = 50 m

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$\star\;{\boxed{\sf{\purple{Area = \dfrac{1}{2} \times (base) \times (height)}}}}$

$:\implies\sf 600 = \dfrac{1}{ \cancel{2}} \times \cancel{50} \times (height)$

$:\implies\sf 600 = 25 \times (height)$

$:\implies\sf Height = 600 \times \dfrac{1}{25}$

$:\implies{\boxed{\frak{\pink{Height = 24\;m}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;length\;of\; altitude\;is\; \bf{24\;m}.}}}$