Question

find the area of the triangle whose sides are 42 CM 34 cm and 20 CM in length hence find the height corresponding to the longest side

Answer 1

ʜᴇʟʟᴏ ᴍᴀᴛᴇ!

Area of triangle by Heron's formula :-

√[ s( s - a )( s - b )( s - c )] where, s = ( a + b + c )/2

s = ( 42 + 34 + 20 ) cm / 2
= ( 96 ) cm / 2 or 48 cm

√[ 48( 48 - 42 )( 48 - 34 )(48-20)

√[ 2³ × 6 × 6 × 2 × 7 × 2² × 7 ]

= ( 2 × 2 × 2 × 6 × 7 ) cm² = 48*7 cm² or 336 cm²

Longest side = 42 cm

Area of ∆ = 1/2 × base × height

336 cm² = 1/2 × 42 cm × h

336 cm² × 2/42 cm = height

8*2 cm = h or height = 16 cm

$\boxed{ Answer : height \: is \: 16 cm }$

Hope it helps

Answer 2

$\sf{GIVEN:-}$

Sides of the triangle is 42cm , 34cm , 20cm.

we know the Heron's formula

area of triangle = $\mathit{\large{\sqrt{s(s - a)(s - b)(s - c)}}}$

where, "s" is the semi perimeter and a,b,c are the sides of the triangle,

So, S = (a + b + c)/2

S = (42 + 34 + 20)/2

S = (96)/2

S = 48cm

Now,

area of triangle = $\mathit{\large{\sqrt{48(48 - 42)(48 - 34)(48 - 20)}}}$

= $\mathit{\large{\sqrt{48 * 6 * 14 * 28}}}$

= $\mathit{\large{\sqrt{(8*6) * 6 * (7 * 2) * (7 * 4)}}}$

= $\mathit{\large{\sqrt{(6*6) * (7*7) * (8*2) * 4}}}$

= $\mathit{\large{\sqrt{36 * 49 * 16 * 4}}}$

= $\mathit{\large{6*7*4*2}}$

= $\mathit{\large{336cm^2}}$

Hence, area of triangle = 336cm²

let the height of the triangle be xcm.
it is given to find the height of the triangle by longest side so, the longest side is 42cm.

now, area of triangle = $\mathit{\large{\frac{1}{2}*42*X = 336}}$

$\mathit{\large{336 = 21X}}$

$\mathit{\large{X = \frac{336}{21}}}$

$\mathit{\large{X = 16cm}}$

$\boxed{\mathit{\large{HENCE, area\:\: of \:\:the\:\: triangle = 336cm^2\;\; and \;\; height \:\: of \:\: the\:\: triangle\:\: = 16cm}}}$