Question

the sides of a triangle are 35cm ,54cm and 61cm.find the area

Answer 1

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


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


$s = \frac{150}{2}$

$s = 75$

$area = \sqrt{s(s - a)(s - b)(s - c)}$

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

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

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

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

$= 420 \sqrt{5}$
$= 420 + 2.236$

$= 422.236 {cm}^{2}$

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Answer 2

$area \: of \: triagle \: \sqrt{s(s - a)(s - b)(s -c)} \\ triangle \: side \: are \: 35cm \: 54cm \: 61cm \\ let \: side \: to \: be \: a. \: b. \: c. \\ s = \frac{a + b + c}{2} \\ s = \frac{35 + 54 + 61}{2} \\ s = 75 \\ now \\ area \: of \: triangle \\ \sqrt{75(75 - 35)(75 - 54)(75 - 61)} \\ \sqrt{75 \times 40 \times 21 \times 14} \\ = 420 \sqrt{5} {cm}^{2}$