Question

The sides of a triangle are 45 cm ,39 cm, 42 cm, find its area​

Answer 1

Answer:

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Let a = 42cm , b = 39 cm , c = 45cm

are three sides of a Triangle.

s = ( a + b + c ) /2

= ( 42 + 39 + 45 ) /2

= 126/2

= 63

s - a = 63 - 42 = 21

s - b = 63 - 39 = 24

s - c = 63 - 45 = 18

Therefore ,

By Heron's formula ,

Area of the triangle ,

∆ = √ s( s - a ) ( s - b ) ( s - c )

= √ 63 × 21 × 24 × 18

= √ 9 × 7 × 7 × 3 × 2 × 2 × 6 × 6 × 3

= √ ( 3×3×3×3)×(7×7)×(6×6)×(2×2)

= 3 × 3 × 7 × 6 × 2

= 756 cm²

Answer 2

Answer:

LENGTHS OF A TRIANGLE :-

a = 45 cm,

b = 39 cm

c = 42 cm

BY USING HERONS FORMULA:-

S = $\dfrac{a + b + c }{2}$

s = $\dfrac{45 + 39 + 42 }{2}$

s = $\dfrac{126}{2}$

s =$\boxed{\sf{63 cm}}$

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

$\implies$ $\sqrt{ 63(63 - 45 ) ( 63 - 39) (63 - 42)}$

$\implies$ $\sqrt{ 63 × 18 × 24 × 21}$

$\implies$ $\sqrt{ 9 × 7 × 6 × 3 × 6 × 4 × 7 × 3}$

$\implies$ $3 × 7 × 6 × 3 × 2$

Area = $\boxed{\sf{756 sq.cm}}$