Question

the perimeter of a triangular field is 240 if two of its sides are 78 metre and 50 find the length of a perpendicular on the side of length 50 metre from opposite vertex using heron's formula
i want the ans asap​

Answer 1

GivEn:

• The perimeter of a triangular field is 240 m.

• Two sides of Triangle are 78 m and 50 m.

To find:

• the length of a perpendicular on the side of length 50 metre from opposite vertex using heron's formula

SoluTion:

GivEn that,

Two sides of Triangle are 78 m and 50 m.

• Let's a = 78 m
• Let's b = 50 m

And the perimeter of a triangular field is 240 m.

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we know that,

Perimeter of a ∆ is = sum of its all sides (a + b + c)

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Therefore,

$:\implies$ 240 = 78 m + 50 m + c

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$:\implies$ c = 240 - (78 + 50)

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$:\implies$ c = 240 - 128

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$:\implies$ c = 112 m

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★ Now, Using Heron's Formula -

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

Semiperimeter,

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

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$:\implies\sf s = \dfrac{78 + 50 + 112}{2}$

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$:\implies\sf s = \cancel{ \dfrac{240}{2}}$

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$:\implies\bf s = 120\;m$

Now,

$:\implies\sf A = \sqrt{s(s - a)(s - b)(s - c)}$

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$:\implies\sf A = \sqrt{120(120 - 78)(120 - 50)(120 - 112)}$

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$:\implies\sf A = \sqrt{120(42)(70)(8)}$

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$:\implies\sf A = \sqrt{2422400}$

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$:\implies{\underline{\boxed{\sf{\pink{A = 1640\;m^2}}}}}\;\bigstar$

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★ Also Area of ∆ can be find by using its base and height,

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

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$:\implies\sf 1640 = \dfrac{1}{2} \times 50 \times H$

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$:\implies\sf H = \dfrac{1640 \times \cancel{2}}{ \cancel{50}}$

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$:\implies\sf H = \cancel{ \dfrac{1640}{25}}$

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$:\implies{\underline{\boxed{\sf{\blue{67.2\;m}}}}}\;\bigstar$

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$\therefore$ Hence, Height of the Triangle is 67.2 m.

Answer 2

Solution :

$\underline{\bf{Given\::}}}$

The perimeter of a triangular field is 240 m, if two of It's sides are 78 m & 50 m.Perpendicular on the side of length 50 m from opposite vertex .

$\underline{\bf{Explanation\::}}}$

Attachment a diagram according to the question;

As we know that formula of the perimeter of triangle;

$\boxed{\bf{Perimter=Side+Side+Side}}}$

A/q

$\longrightarrow\sf{Perimeter=Side+Side+Side}\\\\\longrightarrow\sf{240=78 + 50 + Side}\\\\\longrightarrow\sf{240=128 + Side}\\\\\longrightarrow\sf{Side=240-128}\\\\\longrightarrow\bf{Side=112\:m}$

∴ We have get three side of triangular field;

• A = 78 m, B = 50 m, C = 112 m

Now;

$\underline{\boldsymbol{By\:using\:Heron's\:formula\::}}}$

$\mapsto\sf{Semi-perimeter=\dfrac{Side+Side+Side}{2}} \\\\\\\mapsto\sf{Semi-perimeter = \dfrac{A+B+C}{2} }\\\\\\\mapsto\sf{Semi-perimeter=\dfrac{78m+50m+112m}{2} }\\\\\\\mapsto\sf{Semi-perimeter=\cancel{\dfrac{240}{2} }cm}\\\\\\\mapsto\bf{Semi-perimeter=120\:m}$

&

$\mapsto\sf{Area\:of\:\triangle = \sqrt{S(S-A)(S-B)(S-C)}} \\\\\mapsto\sf{Area\:of\:\triangle = \sqrt{120(120-78)(120-50)(120-112)}}\\\\\mapsto\sf{Area\:of\:\triangle = \sqrt{120(42)(70)(8)}} \\\\\mapsto\sf{Area\:of\:\triangle = \sqrt{2822400} }\\\\\mapsto\bf{Area \:of\:\triangle = 1680\:m^{2}}$

Now,again use formula of an another area of triangle :

$\longrightarrow\sf{Area\:of\:triangle =\dfrac{1}{2} \times Base \times Height }\\ \\ \longrightarrow\sf{ 1680 = 1/2 \times BC \times AD}\\\\\longrightarrow\sf{1680 = 1/\cancel{2} \times \cancel{50} \times AD}\\\\\longrightarrow\sf{1680 = 25 \times AD}\\\\\longrightarrow\sf{AD = \cancel{1680 /25}}\\\\\longrightarrow\bf{AD = 67.2\:m}$

Thus;

The length (height) of the triangular field will be 67.2 m.