Question

find the area of a triangle whose sides are 41 cm ,28 ,15 also find the length of the altitude corresponding to the largest side of the triangle.

Answer 1

Answer i cant answer the question ut i can answer the similer one if it helps plz mark as brainleast

Step-by-step explanation:make a trianlge ABC of 1 side=34cm one side=20 cm n the last side(base of the triangle) of 42cm

now use the heron's formula..according to the herons formula area of a triangle=under root [s(s-1st side of trianlge)(s-2nd side of triangle)(s-3rd side of trianle)].here s denotes the half of perimeter...perimeter is equals to the sum of all the sides of the triangle...that is 34+20+42 and it equals to 96..!!

S is half of the perimeter........here our perimeter is 96.........so its half is 48...!!!

now just put your values in the formula

under root[48(48-34)(48-20)(48-42)]

under root[48*6*14*28]

under root(112896)

=336 cm.

it is given in the question to find the height corresponding to the largest side...here the largest side is 42cm....so it means that we have to take 42cm as our base......now area of triangle is also 1/2(half)*base of the triangle*height of the triangle

A=1/2*B*H

336=1/2*42*H

336=21*H

H=336/21

H=16cm..!!

Answer 2

all sides of ∆ is different so it is heterogeneous ∆.


Area of ∆=
$= \sqrt{s(s - a) \times (s - b) \times ( s - c}$
s= perimeter/2 = (41+28+15)/2 =42
a=41, b=28 & c=15

Area of triangle = √{42×(42-41)×(42-28)×(42-15)}

Area of ∆ = √( 42 ×1 ×14 ×27)

Area ===== =126 cm

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We know

Area of ∆ = 1/2 ×base × height =126

1/2 × 41×h = 126

h =( 126×2)/41

h=6.146 cm