Question

Find the areacof a traingle of sides 14 cm , 15 cm , 16cm . Also , find the length of the perpendicular form the vertex opposite sides of length 15 cm . plz guys answer fast if answer correct i will follow you plz​

Answer 1

Given: Sides of a ∆ are 14 cm, 15 cm and 16 cm.

To find: Length of Perpendicular?

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☯ Let length of perpendicular from the vertex opposite sides of length 15 cm be p.

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⋆ Reference of image is shown in diagram:

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$\setlength{\unitlength}{2.5mm}\begin{picture}(0,0)\linethickness{0.3mm}\qbezier(0,0)(0,0)(8,17)\qbezier(0,0)(0,0)(18,0)\qbezier(18,0)(18,0)(8,17)\put(8,17.8){\sf A}\put( - 1.5, - 1){\sf B}\put(18.5,-1){\sf C}\put(6.5, - 1.5){\sf 15 m}\put(15,8.1){\sf 14 m}\put( - 0,8.1){\sf 16 m}\put(9,8){\sf p}\multiput(8,0)(0,0.5){35}{\line(0,1){0.2}}\end{picture}$

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$\bigstar\:{\underline{\sf{Using\:Heron's\:formula\::}}}$

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

$\sf Where \begin{cases} & \sf{a = \bf{16\;cm}} \\ & \sf{b = \bf{14\;cm}} \\ & \sf{c = \bf{15\;cm}} \end{cases}$

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

$:\implies\sf s = \dfrac{16 + 14 + 15}{2}$

$:\implies\sf s = \cancel{\dfrac{45}{2}}$

$:\implies{\underline{\boxed{\sf{\purple{s = 22.5\;cm}}}}}\;\bigstar$

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$:\implies\sf Area_{\;(triangle)} = \sqrt{22.5(22.5 - 16)(22.5 - 14)(22.5 - 15)}$

$:\implies\sf Area_{\;(triangle)} = \sqrt{22.5 \times 6.5 \times 8.5 \times 7.5}$

$:\implies\sf Area_{\;(triangle)} = \sqrt{9323.43}$

$:\implies{\underline{\boxed{\sf{\pink{Area_{\;(triangle)} = 96.55\:cm^2}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Area\:of\:triangle\;is\; \bf{96.55\:cm^2}.}}}$

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☯ Area of triangle is also given by,

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$\star\;{\boxed{\sf{\pink{Area_{\;(triangle)} = \dfrac{1}{2} \times base \times height}}}}$

$:\implies\sf 96.55 = \dfrac{1}{2} \times 15 \times p$

$:\implies\sf 15 \times p = 96.55 \times 2$

$:\implies\sf 15 \times p = 193.1$

$:\implies\sf p = \cancel{ \dfrac{193.1}{15}}$

$:\implies{\underline{\boxed{\sf{\purple{p = 12.873\:cm\:\:(approx.)}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Length\:of\: perpendicular\;is\: {\textsf{\textbf{12.873\;cm}}}.}}}$