Question

find the aera of given triangle 19m. 17m. 14m.

Answer 1

Given :-

The three sides of the triangle = 19 m, 17 m, 14 m

To be found :-

The area  of the triangles

We could find the area of the triangle using the formula

= ¹/₂ × base × height

We can take any side of the triangle as base but height id not given.

So,

Another method to find the area of the triangle is

$\bf \sqrt{s(s-a)(s-b)(s-c)}$   unit sq.

[ ∴ In which s is the semiperimeter of the triangle and a, b, c are the sides of the triangle ]

Semiperimeter (s) = Perimeter/2

                              = (19 + 17 + 14)/2

                              = (50)/2

                               = 25

So,

\bf \sqrt{s(s-a)(s-b)(s-c)}

= $\bf \sqrt{25(25-19)(25-17)(25-14)}$

= $\bf \sqrt{25(6)(8)(11)}$

= $\bf \sqrt{5 \times 5 \times 2 \times 3 \times 2 \times 2\times 2 \times 11}$

= $\bf 5\times 2\times 2\sqrt{3 \times 11}$

= $\bf 20\sqrt{33}$

Hence the area of the triangle is  $\bf 20\sqrt{33}$ m²

Answer 2

Answer:

$\bold{20 \sqrt{33} \: m^2}$

Step-by-step explanation:

Let a, b, c be the three sides of the triangle. Then, further, let a = 19 m, b = 17 m, and c = 14 m.

Semiperimeter (s) = $\dfrac{a+b+c}{2} = \dfrac{19+17+14}{2} = \dfrac{50}{2} = 25 \: m$

Then, Area of triangle

$= \sqrt{s(s-a)(s-b)(s-c)}\\\\= \sqrt{25(25-19)(25-17)(25-14)}\\\\= \sqrt{5 \times 5 \times 2 \times 3 \times 2 \times 2 \times 2 \times 11}\\\\= 2 \times 2 \times 5 \times \sqrt{33}\\\\= \bold{20 \sqrt{33} \: m^2}$