Question

In tiangle abc sides are 17,25,28.then length of largest altitude

Answer 1

Heya !!

Here's your answer..⬇
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Sides of triangle..,

a = 17
b = 25
c = 28

S = ( a + b + c )/2
S = ( 17 + 25 + 28 )/2
S = 70/2
S = 35

Area of triangle  = √[s(s-a)(s-b)(s-c)]

Area = √[ 35( 35-17 )( 35-25 )( 35-28 ) ]

= √[ 35( 18 )( 10 )( 7 )]
= √[ 35( 1260 )]
= √44100
= 210

Area of ∆ is 210 unit²

Area of triangle = 1/2 × base × altitude

Take base as largest side.

Area = 1/2 × 28 × altitude

210 = 1/2 × 28 × altitude

420 = 28 × altitude

altitude = 420/28

altitude = 15 unit
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Hope it helps..
Thanks :)

Answer 2

Answer :-

Sides of triangle..,

a = 17

b = 25

c = 28

S = ( a + b + c )/2

S = ( 17 + 25 + 28 )/2

S = 70/2

S = 35

Area of triangle  = √[s(s-a)(s-b)(s-c)]

Area = √[ 35( 35-17 )( 35-25 )( 35-28 ) ]

= √[ 35( 18 )( 10 )( 7 )]

= √[ 35( 1260 )]

= √44100

= 210

Area of ∆ is 210 unit²

Area of triangle = 1/2 × base × altitude

Take base as largest side.

Area = 1/2 × 28 × altitude

210 = 1/2 × 28 × altitude

420 = 28 × altitude

altitude = 420/28

altitude = 15 unit