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The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude

(A) $16\sqrt{5}cm$

(B) $10\sqrt{5}cm$

(C) $24\sqrt{5}cm$

(D) 28 $cm$

Answers (1)

$Given\; a= \; 35cm,\; b= 54cm\; ,c= 61cm$

$S= \frac{a+b+c}{2}$

$S= \frac{35+54+61}{2}$

$\Rightarrow S= 75cm$

Using Heron's formula area of triangle $= \; \sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$

$= \; \sqrt{75\left ( 75-35 \right )\left ( 75-54 \right )\left ( 75-61 \right )}$

$= \sqrt{75\times 40\times 21\times 14}$

$= \sqrt{5\times 5\times 3\times 5\times 4\times 2\times 7\times 3\times 7\times 2}= \sqrt{5\times 5\times 5\times 3\times 3\times 2\times 2\times 2\times 2\times 7\times 7}$

$=$ $5\times 3\times 2\times 7\times 2\sqrt{5}$

$= 420\sqrt{5}cm^{2}$

We know that the area of $\bigtriangleup$ ABC $= \frac{1}{2}\times base\times altitude$

$420\sqrt{5}= \frac{1}{2}\times 35\times AD$

$420\sqrt{5}\times 2= 35\times altitude$

$\Rightarrow \frac{420\sqrt{5}\times 2}{35}= \; altitude$

$\Rightarrow \frac{60\sqrt{5}\times 2}{5}= \; altitude$

$\Rightarrow 24\sqrt{5}= \; altitude$

$Hence,altitude\; is \; 24\sqrt{5}cm$

Hence option (C) is correct.

Posted by

infoexpert21

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