Question

side of triangle are in the ratio 10:19:25.perimeter is 540.find area​

Answer 1

Answer:

Step-by-step explanation:

solution :

perimeter = 540 units

ratio = 10:19:25 = 10x:19x:25x

∴ sum of ratios = 10x + 19x + 25x = 540

∴ 54x = 540

∴ x = 540 / 54 = 10

side 1 = 10x = 10 * 10 = 100units

side 2 = 19x = 19 * 10 = 190 units

side 3 = 25x = 25 * 10 = 250 units

by using heron's formula

semi perimeter(s)  = perimeter /2 = 540/2 = 270

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

       = $\sqrt{270(270-100)(270-190)(270-250)\\}$

      = $\sqrt{270(170)(80)(20)}$

      = $\sqrt{5*2*3*3*3*5*2*2*2*2*2*5*2*2}$

      =$240\sqrt{15} units^{2}$

hope it helps :)

Answer 2

Answer:

Perimeter=10x+19x+25x

54x

54x=540

x=10

Sides are 10*10,10*19,10*25

100,190,250

Area

$A = \sqrt{s(s - a)(s - b)(s - c)} \\ where \\ s = \frac{a + b + c}{2}$

s=(100+190+250)/2

s=540/2

s=270

A=√{(270)(270-100)(270-190)(270-250)}

=√(270)(170)(80)(20)

=8570