Question

Find the area of triangle whose sides are 18cm 24cm and 30cm also find the length of altitude corresponding to the largest side of triangle .

Answer 1

a=18
b=24
c=30

s=18+24+30÷2
s=36cm

area=√s(s-a)(s-b)(s-c)
area=√36(36-18)(36-24)(36-30)
area=√36(18)(12)(6)
area=6√2×9×4×3×2×3
area=6×2×2×3×3
area=216cm.sq

area=1/2×b×h
216=1/2×30×h
216=15×h
h=216/15
h=14.4cm

Answer 2

Step-by-step explanation:

GivEn:

Sides of ∆ = 18 cm, 24 cm and 30 cm

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To find:

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

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Solution:

$\frak{Here} \begin{cases} & \sf{a = 18\;cm} \\ & \sf{b = 24\;cm} \\ & \sf{c = 30\;cm} \end{cases}$

✇ Finding semi - perimeter of ∆,

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

$:\implies\sf s = \dfrac{18 + 24 + 30}{2}$

$:\implies\sf s = \dfrac{72}{2}$

$:\implies\bf s = 36\;cm$

✇ Now, Finding area of ∆ using Heron's Formula,

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

$:\implies\sf \sqrt{36(36 - 18)(36 - 24)(36 - 30)}$

$:\implies\sf \sqrt{36 \times 18 \times 12 \times 6}$

$:\implies\sf \sqrt{46656}$

$:\implies{\boxed{\frak{\pink{216\;cm^2}}}}\;\bigstar$

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

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Here,

Largest side of ∆ is = 30 cm

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$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\linethickness{0.3mm}\qbezier(1, 0)(1,0)( 3,3)\qbezier(5,0)(5, 0)(3,3)\qbezier(5,0)(1,0)(1,0)\qbezier(1,0)(1,0)(4,1.5)\put(2.8,3.2){ \sf A }\put(2.5, - 0.3){24 m}}\put(0.5,-0.2){\sf B}\put(4.5,1){30 m}\put(5.1, - 0.3){\sf C}\put(1.1,1.8){18 m}\put(4.2, 1.5){\sf D}\end{picture}$

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

$:\implies\sf 216 = \dfrac{1}{ \cancel{2}} \times \cancel{30} \times (height)$

$:\implies\sf 216 = 15 \times (height)$

$:\implies\sf Height = 216 \times \dfrac{1}{15}$

$:\implies{\boxed{\frak{\pink{Height = 14.4\;cm}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;length\;of\; altitude\;is\; \bf{14.4\;cm}.}}}$