Question

find the area of a triangle whose side are 42cm , 34cm, and 20cm ina length. hence, find the hight corresponding to the longest side

Answer 1

$\bf \large \it{Hey \: User!!!}$

given the sides of the triangle are 42cm, 34cm and 20cm.

let the sides be a, b and c respectively.

s (semi-perimeter) of the triangle = (42+34+20)/2
= 96/2
= 48cm

$\bf \small \: area \: of \: the \: triangle \: by \: herons \: \\ \bf \small formula = \sqrt{s(s - a)(s - b)(s - c)} \\ \bf \small \: = \sqrt{48(48 - 42)(48 - 34)(48 - 20)} \\ \bf \small = \sqrt{48 \times 8 \times 14 \times 28} \\ \bf \small = \sqrt{150528} \\ \bf \small \: = 2 \times 2 \times 2 \times 2 \times 2 \times 7 \sqrt{3} \\ \bf \small = 224 \sqrt{3} {cm}^{2} \\ \bf \small \: = 387 {cm}^{2} (approximately)$

now, we have to find the height of corresponding to the longest side of the triangle.

here, the longest side of the triangle is 42cm.

so the base be 42cm.

therefore 1/2 × b × h = 387cm²
>> 1/2 × 42 × h = 387 cm²
>> 21 × h = 387 cm²
>> h = 387/21
>> h = 18 cm (approximately)

$\bf \large \it{Cheers!!!}$