Question

Sides or triangle are in ratio 12:17:25 and its perimeter is 540cm find its area

Answer 1

Let the sides of triangle be
12x+17x+25x
perimeter of triangle = sum its 3 sides
540 cm = 12x +17x +25x
540 =54 x
540/54= x
10 =x
Sides of triangle will be 12x=12×10=120cm (base)
17x=17×10=170 cm (height)
25x=25×10=250 cm (hypotenuse longest side)

now area of triangle =1/2× base ×height
=1/2× 120 ×170
=10200cm²

Answer 2

Explanation:

Given:

The ratio of the sides if a triangle is 12:17:25

The perimeter of the triangle is 540 cm

To find:

The area of the triangle

So,

Let the sides of the triangle be

a = 12x cm

b = 17x cm

c = 25x cm

According to question,

The perimeter of the triangle = 540

$⇒ a + b + c = 540 \\ </h3><h3></h3><h3>⇒ 12x + 17x + 25x = 540 \\ </h3><h3></h3><h3>⇒ 54x = 540 \\ </h3><h3></h3><h3>⇒ x = 540 ÷ 54$

[By transporting 54 to RHS]

$⇒ x = 10$

Thus,

The value of x is 10

Now the sides of the triangle are

$a = 12x = 12 * 10 = 120 cm \\ </h3><h3>b = 17x = 17 * 10 = 170 cm \\ </h3><h3>c = 25x = 25 * 10 = 250 cm$

We can find the area of the triangle bye Heron's formula

$S = semiperimeter = Periemter/2$

$S = 540/2 = 270$

Now,

$=\sqrt{S(S-a)(S-b)(S-c)}= </h3><h3>$

units sq.

$=\sqrt{270(270-120)(270-170)(270-250)}$

$=\sqrt{270(150)(100)(20)}$

$=\sqrt{3*3*3*10*(3*5*10)(10*10)(2*10)}$

$=\sqrt{\underline{3*3}*\underline{3*3}*\underline{10*10}*\underline{10*10}*10*2*5}</p><p>$

$=\sqrt{\underline{3*3}*\underline{3*3}*\underline{10*10}*\underline{10*10}*\underline{10*10} }$

$= 3 * 3 * 10 * 10 * 10</p><p>$

= 9000 cm sq.

$hope \: its \: help \: u$