Question

If the sides of triangle are 20cm, 24cm and 28cm, find the length of the altitude using longest side as base

Answer 1

Step-by-step explanation:

=s+s+s

=(20+24+28)

=(44+28)

=(72)answers

Answer 2

Given sides of triangle are 20cm, 24cm and 28cm.

We have to find, length of the altitude.

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⋆ Reference of image is shown in diagram

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$\setlength{\unitlength}{1.6mm}\begin{picture}(50,20)\linethickness{0.1mm}\put(-3,-3){\line(1,1){20}}\put(36.6,-2.8){\line(-1,1){19.8}}\put(-3,-3){\line(1,0){39.5}}\put(30,5){20 cm}\put(-1.5,5){24 cm}\put(3,5){}\put(15,-5){28 cm}\put(15.5,17.5){A}\put(-4,-5){B}\put(35,-5){C}\end{picture}$

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$\dag\;{\underline{\frak{We\;know\;that,}}}$

☯ Area of ∆ can be find by using Heron's Formula,

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

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$\frak Here \begin{cases} & \text{a = 20} \\ & \text{b = 24 } \\ & \text{c = 28} \end{cases}$

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$:\implies\sf s = semi - perimeter$

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

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$:\implies\sf s = \dfrac{20 + 24 + 28}{2}$

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

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

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$\dag\;{\underline{\frak{Now,\;Putting\;values\;:}}}$

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$:\implies\sf \sqrt{36(36 - 20)(36 - 24)(36 - 28)}$

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$:\implies\sf \sqrt{36(8)(12)(16)}$

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$:\implies\sf \sqrt{36 \times 8 \times 12 \times 16}$

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$:\implies\sf \sqrt{55296}$

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$:\implies{\underline{\boxed{\sf{\purple{235.15\;cm^2}}}}}\;\bigstar$

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$\dag\;{\underline{\frak{We\;know\;that,}}}$

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☯ Area of ∆ also equal to,

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$\star\;{\boxed{\sf{\pink{Area = \frac{1}{2} \times bc \times sin\;A}}}}$

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$\frak Here \begin{cases} & \text{Area = 235.15 \sf cm^2 } \\ & \text{b = 24 cm} \\ & \text{c = 28 cm} \end{cases}$

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$\dag\;{\underline{\frak{Now,\;Putting\;values\;:}}}$

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$:\implies\sf 235.15 = \frac{1}{2} \times 24 \times 28 \times \frac{h}{a}$

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$:\implies\sf 235.15 = \frac{1}{ \cancel{2}} \times \cancel{24} \times 28 \times \frac{h}{a}$

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$:\implies\sf 235.15 = 12 \times 28 \times \frac{h}{20}$

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$:\implies\sf 235.15 \times 20 = 336 \times h$

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$:\implies\sf 4703 = 336 \times h$

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$:\implies\sf h = \cancel{ \dfrac{4703}{336}}$

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$:\implies{\underline{\boxed{\sf{\pink{h = 13.997\;cm}}}}}\;\bigstar$

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$\therefore$ Height or Length of Altitude of given triangle is 13.997 cm.