Question

the sides of a triangle are in the ratio 14:20:25 and its perimeter is 590cm. find its area​

Answer 1

let us consider the common ratio between the sides of the triangle be "a"

therefore the sides are 14a , 20a and 25a

perimeter = 590 cm

$14a + 20a + 25a = 590cm$

$59a = 590$

$\: a \: = 10$

now the sides of the triangles are 140cm ,200cm ,250cm

so,the semi perimeter of the triangle (s)

$= \frac{590}{2}$

$= 295cm$

Using Heron's formula for area of the triangle

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

= √295(295-140)(295-200)(295-250)

$= \sqrt{295 \times 155 \times 95 \times 45}$

$= \sqrt{195474375}$

$= 13981.21 {cm}^{2}$

Answer 2

Answer:

The ratio of the sides of the triangle is given as 14: 20: 25Let us consider the common ratio

between the sides of the triangle be “a”∴ The sides are 14a, 20a and 25a

It is also given that the perimeter of the triangle = 590 cm

12a + 17a + 25a = 590 => 59a = 590So, a = 10Now, the sides of the triangle are 140 cm, 200 cm, 250 cm.So, the

semi perimeter of the triangle (s) = 590/2 = 295 cm

Using Heron’s formula for Area of the triangle=

√s(s−a)(s−b)(s−c)

√s(s-a)(s-b)(s-c)}= 295(295−140)(295−200)(295−250)

√295(295−140)(295−200)(295−250)

√295(295-140)(295-200)(295-250)}=

√295×155×95×45

√195,474,375\

= 13981.21cm2